OP Malhotra Class 11 Maths Solutions Chapter 24 Ellipse Exercise 24

Question 1. Find the eccentricity of the ellipse of which the major axis is double the minor axis.
Answer: Let \(a\) be the length of the semi-major axis and \(b\) be the length of the semi-minor axis of the ellipse. According to the given condition, the major axis is double the minor axis, so \(a = 2b\). We know the standard relationship between \(a\), \(b\), and eccentricity \(e\) for an ellipse: \( b^2 = a^2 (1 - e^2) \). Now, substitute \(a = 2b\) into this equation: \( b^2 = (2b)^2 (1 - e^2) \) \( b^2 = 4b^2 (1 - e^2) \) Since \(b\) is a length, \(b \ne 0\). We can divide both sides by \(b^2\): \( 1 = 4 (1 - e^2) \) \( \frac{1}{4} = 1 - e^2 \) \( e^2 = 1 - \frac{1}{4} \) \( e^2 = \frac{3}{4} \)
\( \implies \) \( e = \frac{\sqrt{3}}{2} \) (Since eccentricity \(e\) must be greater than 0) Thus, the required eccentricity of the ellipse is \( \frac{\sqrt{3}}{2} \). The eccentricity measures how "stretched" an ellipse is; a value close to 0 is nearly a circle, while a value close to 1 is very flat.In simple words: If the long side of an ellipse is twice the short side, its flatness (eccentricity) is calculated to be root 3 divided by 2.

🎯 Exam Tip: Always start by clearly defining the semi-major axis \(a\) and semi-minor axis \(b\), and use the fundamental relation \( b^2 = a^2 (1 - e^2) \). Remember that eccentricity \(e\) must always be a positive value between 0 and 1.

Question 2. If the minor axis of an ellipse is equal to the distance between its foci, prove that its eccentricity is \( \frac{1}{\sqrt{2}} \).
Answer: Let \(a\) and \(b\) be the lengths of the semi-major and semi-minor axes, respectively, and let \(e\) be the eccentricity of the ellipse. The length of the minor axis is \(2b\). The distance between its foci is \(2ae\). According to the given condition, these two lengths are equal: \( 2b = 2ae \) Dividing both sides by 2, we get: \( b = ae \) We know the fundamental relationship for an ellipse: \( b^2 = a^2 (1 - e^2) \). Substitute \(b = ae\) into this equation: \( (ae)^2 = a^2 (1 - e^2) \) \( a^2e^2 = a^2 (1 - e^2) \) Since \(a\) is the semi-major axis length, \(a \ne 0\). We can divide both sides by \(a^2\): \( e^2 = 1 - e^2 \) Add \(e^2\) to both sides: \( 2e^2 = 1 \) \( e^2 = \frac{1}{2} \)
\( \implies \) \( e = \frac{1}{\sqrt{2}} \) (Since eccentricity \(e\) must be positive) Thus, the eccentricity of the ellipse is \( \frac{1}{\sqrt{2}} \). When the eccentricity is \(1/\sqrt{2}\), it means the ellipse is a specific shape where the minor axis length is exactly the same as the distance between its two focal points.In simple words: If the short width of an ellipse is the same as the distance between its two special points (foci), then its flatness value (eccentricity) is exactly 1 divided by root 2.

🎯 Exam Tip: Clearly define the terms "minor axis" (\(2b\)) and "distance between foci" (\(2ae\)) and set up the initial equality correctly. Using the fundamental equation \(b^2 = a^2(1-e^2)\) is key to solving this type of problem.

Question 3. Find the latus rectum and eccentricity of the ellipse whose semi-axes are 5 and 4.
Answer: Let the lengths of the semi-major axis be \(a\) and the semi-minor axis be \(b\). Given the semi-axes are 5 and 4. We always take the larger value as the semi-major axis, so \(a = 5\) and \(b = 4\). First, let's find the eccentricity \(e\). We use the relationship: \( b^2 = a^2 (1 - e^2) \). Substitute the values of \(a\) and \(b\): \( 4^2 = 5^2 (1 - e^2) \) \( 16 = 25 (1 - e^2) \) Divide by 25: \( \frac{16}{25} = 1 - e^2 \)
\( \implies \) \( e^2 = 1 - \frac{16}{25} \) \( e^2 = \frac{25 - 16}{25} \) \( e^2 = \frac{9}{25} \)
\( \implies \) \( e = \frac{3}{5} \) (Since eccentricity \(e\) must be positive) Next, let's find the length of the latus rectum (LR). The formula for the latus rectum is: \( LR = \frac{2b^2}{a} \). Substitute the values of \(a\) and \(b\): \( LR = \frac{2 \times (4)^2}{5} \) \( LR = \frac{2 \times 16}{5} \) \( LR = \frac{32}{5} \) So, the eccentricity of the ellipse is \( \frac{3}{5} \) and the length of the latus rectum is \( \frac{32}{5} \). The latus rectum is a chord that passes through a focus and is perpendicular to the major axis, giving another important measure of the ellipse's shape.In simple words: For an ellipse with semi-axes of 5 and 4, its flatness (eccentricity) is 3/5, and the length of its special chord (latus rectum) is 32/5.

🎯 Exam Tip: Always identify \(a\) as the semi-major axis and \(b\) as the semi-minor axis correctly (i.e., \(a > b\)). Remember the formulas for eccentricity \(e\) and latus rectum \(LR\) to avoid losing marks.

Question 4. Find the eccentricity of the ellipse whose latus rectum is (i) half its major axis, (ii) half its minor axis.
Answer: Let \(a\) be the length of the semi-major axis, \(b\) be the length of the semi-minor axis, and \(e\) be the eccentricity of the ellipse. The formula for the length of the latus rectum (LR) is \( \frac{2b^2}{a} \). (i) According to the given condition, the latus rectum is half its major axis. Major axis length is \(2a\). So, \( LR = \frac{1}{2} (2a) = a \). Now, equate this with the formula for the latus rectum: \( \frac{2b^2}{a} = a \)
\( \implies \) \( 2b^2 = a^2 \) We also know the fundamental relationship: \( b^2 = a^2 (1 - e^2) \). Substitute \( b^2 = \frac{a^2}{2} \) into this equation: \( \frac{a^2}{2} = a^2 (1 - e^2) \) Since \(a \ne 0\), divide both sides by \(a^2\): \( \frac{1}{2} = 1 - e^2 \)
\( \implies \) \( e^2 = 1 - \frac{1}{2} \) \( e^2 = \frac{1}{2} \)
\( \implies \) \( e = \frac{1}{\sqrt{2}} \) (Since \(e > 0\)) (ii) According to the given condition, the latus rectum is half its minor axis. Minor axis length is \(2b\). So, \( LR = \frac{1}{2} (2b) = b \). Now, equate this with the formula for the latus rectum: \( \frac{2b^2}{a} = b \) Since \(b \ne 0\), divide both sides by \(b\): \( \frac{2b}{a} = 1 \)
\( \implies \) \( 2b = a \) We use the fundamental relationship: \( b^2 = a^2 (1 - e^2) \). Substitute \( a = 2b \) into this equation: \( b^2 = (2b)^2 (1 - e^2) \) \( b^2 = 4b^2 (1 - e^2) \) Since \(b \ne 0\), divide both sides by \(b^2\): \( 1 = 4 (1 - e^2) \) \( \frac{1}{4} = 1 - e^2 \)
\( \implies \) \( e^2 = 1 - \frac{1}{4} \) \( e^2 = \frac{3}{4} \)
\( \implies \) \( e = \frac{\sqrt{3}}{2} \) (Since \(e > 0\)) The latus rectum is an important measurement for the ellipse as it provides insight into its curvature near the foci.In simple words: (i) If the special chord (latus rectum) is half the long side, the ellipse's flatness is 1 over root 2. (ii) If the special chord is half the short side, its flatness is root 3 over 2.

🎯 Exam Tip: Be careful to distinguish between "major axis" (\(2a\)) and "minor axis" (\(2b\)) when setting up the initial equation for the latus rectum. Use the given condition and the eccentricity formula to find \(e\).

Question 5. If the eccentricity is zero, prove that the ellipse becomes a circle.
Answer: Let \(a\) be the length of the semi-major axis, \(b\) be the length of the semi-minor axis, and \(e\) be the eccentricity of the ellipse. The fundamental relationship connecting these values is: \( b^2 = a^2 (1 - e^2) \). We are given that the eccentricity \(e = 0\). Substitute \(e = 0\) into the relationship: \( b^2 = a^2 (1 - 0^2) \) \( b^2 = a^2 (1 - 0) \) \( b^2 = a^2 \) Since \(a\) and \(b\) represent lengths, they must be positive. Therefore, if \(b^2 = a^2\), it implies that \( b = a \). Now, consider the standard equation of an ellipse centered at the origin: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) Since we found that \(b = a\), we can replace \(b^2\) with \(a^2\) in the ellipse equation: \( \frac{x^2}{a^2} + \frac{y^2}{a^2} = 1 \) Multiply the entire equation by \(a^2\) to clear the denominators: \( x^2 + y^2 = a^2 \) This is the standard equation of a circle centered at the origin with a radius of \(a\). Thus, when the eccentricity of an ellipse is zero, its semi-major and semi-minor axes become equal, causing the ellipse to transform into a circle. A circle is a special type of ellipse where both foci coincide at the center.In simple words: When an ellipse has zero flatness (eccentricity), its long and short sides become equal. This changes the ellipse into a perfect circle.

🎯 Exam Tip: This proof relies on the key relationship \( b^2 = a^2 (1 - e^2) \) and the standard equation of an ellipse. Clearly show how substituting \(e=0\) leads to \(a=b\) and then to the circle equation.

Question 6. Find the equation to the ellipse with axes as the axes of coordinates.
(i) major axis = 6, minor axis = 4 ;
(ii) which passes through the points (-3, 1) and (2, -2) ;
(iii) axes are 10 and 8 and the major axis along
(a) the axis of x,
(b) the axis of y ;
(iv) major axis \( \frac{9}{2} \) and eccentricity \( \frac{1}{\sqrt{3}} \), where the major axis is the horizontal axis ;
(v) latus rectum is 5 and eccentricity \( \frac{2}{3} \),
(vi) foci are (\( \pm 4, 0 \)) and e = \( \frac{1}{3} \);
(vii) distance between the foci is 10 and its latus rectum is 15 ;
(viii) distance of the focus from the corresponding directrix is 9 and eccentricity is \( \frac{4}{5} \);
(ix) the minor axis is equal to the distance between the foci, and the latus rectum is 10.
Answer: Let the equation of the ellipse be \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) for a horizontal major axis, or \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) for a vertical major axis, where \(a\) is the semi-major axis and \(b\) is the semi-minor axis (\(a > b\)). (i) Given major axis \(2a = 6 \implies a = 3\). Given minor axis \(2b = 4 \implies b = 2\). Assuming the major axis is along the x-axis, the equation of the ellipse is: \( \frac{x^2}{3^2} + \frac{y^2}{2^2} = 1 \) \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \) To clear the denominators, multiply by the LCM of 9 and 4 (which is 36): \( 4x^2 + 9y^2 = 36 \) (ii) Let the equation of the ellipse be \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). The ellipse passes through the points (-3, 1) and (2, -2). Substitute (-3, 1): \( \frac{(-3)^2}{a^2} + \frac{1^2}{b^2} = 1 \) \( \frac{9}{a^2} + \frac{1}{b^2} = 1 \) ... (Equation 2) Substitute (2, -2): \( \frac{2^2}{a^2} + \frac{(-2)^2}{b^2} = 1 \) \( \frac{4}{a^2} + \frac{4}{b^2} = 1 \) ... (Equation 3) From Equation 3, divide by 4: \( \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4} \) ... (Equation 4) Subtract Equation 4 from Equation 2: \( \left(\frac{9}{a^2} + \frac{1}{b^2}\right) - \left(\frac{1}{a^2} + \frac{1}{b^2}\right) = 1 - \frac{1}{4} \) \( \frac{8}{a^2} = \frac{3}{4} \)
\( \implies \) \( 3a^2 = 32 \)
\( \implies \) \( a^2 = \frac{32}{3} \) Substitute the value of \(a^2\) into Equation 4: \( \frac{1}{32/3} + \frac{1}{b^2} = \frac{1}{4} \) \( \frac{3}{32} + \frac{1}{b^2} = \frac{1}{4} \) \( \frac{1}{b^2} = \frac{1}{4} - \frac{3}{32} \) \( \frac{1}{b^2} = \frac{8 - 3}{32} \) \( \frac{1}{b^2} = \frac{5}{32} \)
\( \implies \) \( b^2 = \frac{32}{5} \) Now, substitute \(a^2\) and \(b^2\) back into the general equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \): \( \frac{x^2}{32/3} + \frac{y^2}{32/5} = 1 \) \( \frac{3x^2}{32} + \frac{5y^2}{32} = 1 \) To clear the denominator, multiply by 32: \( 3x^2 + 5y^2 = 32 \) (iii) Given axes lengths are 10 and 8. So, \(2a = 10 \implies a = 5\) and \(2b = 8 \implies b = 4\). (a) Major axis along the x-axis: The equation of the ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Substitute \(a=5\) and \(b=4\): \( \frac{x^2}{5^2} + \frac{y^2}{4^2} = 1 \) \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \) To clear the denominators, multiply by the LCM of 25 and 16 (which is 400): \( 16x^2 + 25y^2 = 400 \) (b) Major axis along the y-axis: The equation of the ellipse is \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \). Substitute \(a=5\) and \(b=4\): \( \frac{x^2}{4^2} + \frac{y^2}{5^2} = 1 \) \( \frac{x^2}{16} + \frac{y^2}{25} = 1 \) To clear the denominators, multiply by the LCM of 16 and 25 (which is 400): \( 25x^2 + 16y^2 = 400 \) (iv) Given major axis \(2a = \frac{9}{2} \implies a = \frac{9}{4}\). Given eccentricity \(e = \frac{1}{\sqrt{3}}\). Since the major axis is horizontal, the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is used. First, find \(b^2\) using the relationship \( b^2 = a^2 (1 - e^2) \): \( b^2 = \left(\frac{9}{4}\right)^2 \left(1 - \left(\frac{1}{\sqrt{3}}\right)^2\right) \) \( b^2 = \frac{81}{16} \left(1 - \frac{1}{3}\right) \) \( b^2 = \frac{81}{16} \left(\frac{2}{3}\right) \) \( b^2 = \frac{27}{8} \) Now, substitute \(a^2 = \left(\frac{9}{4}\right)^2 = \frac{81}{16}\) and \(b^2 = \frac{27}{8}\) into the ellipse equation: \( \frac{x^2}{81/16} + \frac{y^2}{27/8} = 1 \) \( \frac{16x^2}{81} + \frac{8y^2}{27} = 1 \) To clear the denominators, multiply by the LCM of 81 and 27 (which is 81): \( 16x^2 + 8y^2 \left(\frac{81}{27}\right) = 81 \) \( 16x^2 + 8y^2 (3) = 81 \) \( 16x^2 + 24y^2 = 81 \) (v) Given latus rectum \( \frac{2b^2}{a} = 5 \implies b^2 = \frac{5a}{2} \). Given eccentricity \(e = \frac{2}{3}\). Use the relationship \( b^2 = a^2 (1 - e^2) \): \( \frac{5a}{2} = a^2 \left(1 - \left(\frac{2}{3}\right)^2\right) \) \( \frac{5a}{2} = a^2 \left(1 - \frac{4}{9}\right) \) \( \frac{5a}{2} = a^2 \left(\frac{5}{9}\right) \) Since \(a \ne 0\), divide by \(a\) and multiply by \( \frac{9}{5} \): \( \frac{5}{2} = a \left(\frac{5}{9}\right) \)
\( \implies \) \( a = \frac{5}{2} \times \frac{9}{5} \) \( a = \frac{9}{2} \) Now find \(b^2\): \( b^2 = \frac{5a}{2} = \frac{5}{2} \times \frac{9}{2} = \frac{45}{4} \) Substitute \(a^2 = \left(\frac{9}{2}\right)^2 = \frac{81}{4}\) and \(b^2 = \frac{45}{4}\) into the standard ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \): \( \frac{x^2}{81/4} + \frac{y^2}{45/4} = 1 \) \( \frac{4x^2}{81} + \frac{4y^2}{45} = 1 \) To clear the denominators, multiply by the LCM of 81 and 45 (which is 405): \( \frac{4x^2 \times 405}{81} + \frac{4y^2 \times 405}{45} = 405 \) \( 4x^2 \times 5 + 4y^2 \times 9 = 405 \) \( 20x^2 + 36y^2 = 405 \) (vi) Given foci are (\( \pm 4, 0 \)). This means the major axis is along the x-axis, and \(ae = 4\). Given eccentricity \(e = \frac{1}{3}\). Substitute \(e\) into \(ae = 4\): \( a \left(\frac{1}{3}\right) = 4 \)
\( \implies \) \( a = 12 \) Now, find \(b^2\) using the relationship \( b^2 = a^2 (1 - e^2) \): \( b^2 = 12^2 \left(1 - \left(\frac{1}{3}\right)^2\right) \) \( b^2 = 144 \left(1 - \frac{1}{9}\right) \) \( b^2 = 144 \left(\frac{8}{9}\right) \) \( b^2 = 16 \times 8 \) \( b^2 = 128 \) Substitute \(a^2 = 12^2 = 144\) and \(b^2 = 128\) into the standard ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \): \( \frac{x^2}{144} + \frac{y^2}{128} = 1 \) To clear the denominators, multiply by the LCM of 144 and 128 (which is 1152): \( \frac{x^2 \times 1152}{144} + \frac{y^2 \times 1152}{128} = 1152 \) \( 8x^2 + 9y^2 = 1152 \) (vii) Given distance between foci \(2ae = 10 \implies ae = 5\). Given latus rectum \( \frac{2b^2}{a} = 15 \implies b^2 = \frac{15a}{2} \). Use the relationship \( b^2 = a^2 (1 - e^2) \). We can rewrite this as \( b^2 = a^2 - a^2e^2 = a^2 - (ae)^2 \). Substitute \(ae = 5\) and \(b^2 = \frac{15a}{2}\): \( \frac{15a}{2} = a^2 - 5^2 \) \( \frac{15a}{2} = a^2 - 25 \) Multiply by 2: \( 15a = 2a^2 - 50 \) Rearrange into a quadratic equation: \( 2a^2 - 15a - 50 = 0 \) Factor the quadratic: \( 2a^2 - 20a + 5a - 50 = 0 \) \( 2a(a - 10) + 5(a - 10) = 0 \) \( (a - 10)(2a + 5) = 0 \) This gives two possible values for \(a\): \(a = 10\) or \(a = -\frac{5}{2}\). Since \(a\) is a length, \(a\) must be positive, so \(a = 10\). Now find \(b^2\): \( b^2 = \frac{15a}{2} = \frac{15 \times 10}{2} = 15 \times 5 = 75 \) Substitute \(a^2 = 10^2 = 100\) and \(b^2 = 75\) into the standard ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \): \( \frac{x^2}{100} + \frac{y^2}{75} = 1 \) To clear the denominators, multiply by the LCM of 100 and 75 (which is 300): \( \frac{x^2 \times 300}{100} + \frac{y^2 \times 300}{75} = 300 \) \( 3x^2 + 4y^2 = 300 \) (viii) Given distance of focus from corresponding directrix is 9. This distance is given by \( \frac{a}{e} - ae = 9 \). Given eccentricity \(e = \frac{4}{5}\). Substitute \(e\) into the distance equation: \( \frac{a}{4/5} - a \left(\frac{4}{5}\right) = 9 \) \( \frac{5a}{4} - \frac{4a}{5} = 9 \) Find a common denominator (20) for the left side: \( \frac{25a - 16a}{20} = 9 \) \( \frac{9a}{20} = 9 \)
\( \implies \) \( 9a = 9 \times 20 \)
\( \implies \) \( a = 20 \) Now, find \(b^2\) using the relationship \( b^2 = a^2 (1 - e^2) \): \( b^2 = 20^2 \left(1 - \left(\frac{4}{5}\right)^2\right) \) \( b^2 = 400 \left(1 - \frac{16}{25}\right) \) \( b^2 = 400 \left(\frac{25 - 16}{25}\right) \) \( b^2 = 400 \left(\frac{9}{25}\right) \) \( b^2 = 16 \times 9 \) \( b^2 = 144 \) Substitute \(a^2 = 20^2 = 400\) and \(b^2 = 144\) into the standard ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \): \( \frac{x^2}{400} + \frac{y^2}{144} = 1 \) (ix) Given minor axis is equal to the distance between the foci. Minor axis is \(2b\). Distance between foci is \(2ae\). So, \( 2b = 2ae \implies b = ae \). Given latus rectum is 10: \( \frac{2b^2}{a} = 10 \implies b^2 = 5a \). From \(b = ae\), we have \( e = \frac{b}{a} \). Substitute this into the formula \( b^2 = a^2 (1 - e^2) \): \( b^2 = a^2 \left(1 - \left(\frac{b}{a}\right)^2\right) \) \( b^2 = a^2 \left(1 - \frac{b^2}{a^2}\right) \) \( b^2 = a^2 - b^2 \)
\( \implies \) \( 2b^2 = a^2 \) Now we have two equations: \( b^2 = 5a \) and \( 2b^2 = a^2 \). Substitute \(a^2 = 2b^2\) into \(b^2=5a\). Actually, it is easier to solve for 'a' from \(b^2=5a\) as \(a = b^2/5\). Substitute \(a = \frac{b^2}{5}\) into \(2b^2 = a^2\): \( 2b^2 = \left(\frac{b^2}{5}\right)^2 \) \( 2b^2 = \frac{b^4}{25} \) Since \(b \ne 0\), we can divide by \(b^2\): \( 2 = \frac{b^2}{25} \)
\( \implies \) \( b^2 = 2 \times 25 \) \( b^2 = 50 \) Now find \(a\): \( a = \frac{b^2}{5} = \frac{50}{5} = 10 \) Substitute \(a^2 = 10^2 = 100\) and \(b^2 = 50\) into the standard ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \): \( \frac{x^2}{100} + \frac{y^2}{50} = 1 \) To clear the denominators, multiply by the LCM of 100 and 50 (which is 100): \( \frac{x^2 \times 100}{100} + \frac{y^2 \times 100}{50} = 100 \) \( x^2 + 2y^2 = 100 \) Finding the equation of an ellipse involves using given properties to determine the semi-major axis, semi-minor axis, and orientation before assembling the final equation.In simple words: We used different clues like axis lengths, points the ellipse passes through, or special measurements like latus rectum and foci to find the unique equation for each ellipse. Each clue helps us find the values of 'a' and 'b' and how the ellipse is placed.

🎯 Exam Tip: For problems with multiple conditions, first write down all given information in terms of \(a, b, e\), and \(ae\). Then use the fundamental relations \(b^2 = a^2(1-e^2)\) and \(LR = \frac{2b^2}{a}\) to form a system of equations. Always check if the major axis is horizontal or vertical to use the correct equation form.

Question 7. Find the equation of the ellipse whose centre is at (-2, 3) and whose semi-axes are 3 and 2, when the major axis is
(i) parallel to the axes of x ;
(ii) parallel to the axis of y.
Answer: The center of the ellipse is given as \((h, k) = (-2, 3)\). The semi-axes are 3 and 2. This means \(a\) (semi-major axis) is 3 and \(b\) (semi-minor axis) is 2, since \(a > b\). (i) When the major axis is parallel to the x-axis: The general equation of an ellipse with center \((h, k)\) and major axis parallel to the x-axis is: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) Substitute \(h=-2\), \(k=3\), \(a=3\), and \(b=2\): \( \frac{(x - (-2))^2}{3^2} + \frac{(y - 3)^2}{2^2} = 1 \) \( \frac{(x + 2)^2}{9} + \frac{(y - 3)^2}{4} = 1 \) To clear the denominators, multiply the entire equation by the LCM of 9 and 4 (which is 36): \( 4(x + 2)^2 + 9(y - 3)^2 = 36 \) Expand the squared terms: \( 4(x^2 + 4x + 4) + 9(y^2 - 6y + 9) = 36 \) \( 4x^2 + 16x + 16 + 9y^2 - 54y + 81 = 36 \) Rearrange the terms to get the general form of the equation: \( 4x^2 + 9y^2 + 16x - 54y + 97 - 36 = 0 \) \( 4x^2 + 9y^2 + 16x - 54y + 61 = 0 \) (ii) When the major axis is parallel to the y-axis: The general equation of an ellipse with center \((h, k)\) and major axis parallel to the y-axis is: \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \) Substitute \(h=-2\), \(k=3\), \(a=3\), and \(b=2\): \( \frac{(x - (-2))^2}{2^2} + \frac{(y - 3)^2}{3^2} = 1 \) \( \frac{(x + 2)^2}{4} + \frac{(y - 3)^2}{9} = 1 \) To clear the denominators, multiply the entire equation by the LCM of 4 and 9 (which is 36): \( 9(x + 2)^2 + 4(y - 3)^2 = 36 \) Expand the squared terms: \( 9(x^2 + 4x + 4) + 4(y^2 - 6y + 9) = 36 \) \( 9x^2 + 36x + 36 + 4y^2 - 24y + 36 = 36 \) Rearrange the terms to get the general form of the equation: \( 9x^2 + 4y^2 + 36x - 24y + 36 = 0 \) The orientation of the major axis significantly changes the form of the ellipse equation.In simple words: We found the ellipse's equation when its center is at (-2, 3) and its long and short halves are 3 and 2. The equation changes depending on whether the long side runs left-right (x-axis) or up-down (y-axis).

🎯 Exam Tip: Always correctly identify the center \((h, k)\), semi-major axis \(a\), and semi-minor axis \(b\). The orientation of the major axis determines whether \(a^2\) is under \((x-h)^2\) or \((y-k)^2\). Remember to expand and simplify the equation for the final answer.

Question 8. Find the equation of the ellipse with its centre at (4, -1), focus at (1, -1), and passing through (8, 0).
Answer: Let the center of the ellipse be \(C = (4, -1)\) and one focus be \(S = (1, -1)\). Since the y-coordinates of the center and the focus are the same (-1), the major axis of the ellipse is parallel to the x-axis. The distance from the center to a focus is \(ae\). \( ae = CS = \sqrt{(4-1)^2 + (-1 - (-1))^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3 \). So, \(ae = 3\). The general equation of an ellipse with its center \((h, k)\) and major axis parallel to the x-axis is: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) Substitute \(h=4\) and \(k=-1\): \( \frac{(x-4)^2}{a^2} + \frac{(y-(-1))^2}{b^2} = 1 \) \( \frac{(x-4)^2}{a^2} + \frac{(y+1)^2}{b^2} = 1 \) ... (Equation 1) We know the relationship \( b^2 = a^2 (1 - e^2) \), which can also be written as \( b^2 = a^2 - (ae)^2 \). Substitute \(ae = 3\): \( b^2 = a^2 - 3^2 \) \( b^2 = a^2 - 9 \) Substitute this expression for \(b^2\) into Equation 1: \( \frac{(x-4)^2}{a^2} + \frac{(y+1)^2}{a^2-9} = 1 \) ... (Equation 2) The ellipse passes through the point (8, 0). Substitute \(x=8\) and \(y=0\) into Equation 2: \( \frac{(8-4)^2}{a^2} + \frac{(0+1)^2}{a^2-9} = 1 \) \( \frac{4^2}{a^2} + \frac{1^2}{a^2-9} = 1 \) \( \frac{16}{a^2} + \frac{1}{a^2-9} = 1 \) Multiply by \(a^2(a^2-9)\) to clear denominators: \( 16(a^2-9) + a^2 = a^2(a^2-9) \) \( 16a^2 - 144 + a^2 = a^4 - 9a^2 \) \( 17a^2 - 144 = a^4 - 9a^2 \) Rearrange into a quadratic equation in terms of \(a^2\): \( a^4 - 9a^2 - 17a^2 + 144 = 0 \) \( a^4 - 26a^2 + 144 = 0 \) Let \(A = a^2\). The equation becomes \( A^2 - 26A + 144 = 0 \). Factor the quadratic equation: \( (A - 18)(A - 8) = 0 \) So, \( A = 18 \) or \( A = 8 \). This means \( a^2 = 18 \) or \( a^2 = 8 \). If \( a^2 = 8 \), then \( b^2 = a^2 - 9 = 8 - 9 = -1 \). A squared length cannot be negative, so \(a^2 = 8\) is not a valid solution. Therefore, \( a^2 = 18 \). Now find \(b^2\): \( b^2 = a^2 - 9 = 18 - 9 = 9 \). Substitute \(a^2 = 18\) and \(b^2 = 9\) into Equation 1: \( \frac{(x-4)^2}{18} + \frac{(y+1)^2}{9} = 1 \) To clear the denominators, multiply by the LCM of 18 and 9 (which is 18): \( (x-4)^2 + 2(y+1)^2 = 18 \) Expand and simplify: \( (x^2 - 8x + 16) + 2(y^2 + 2y + 1) = 18 \) \( x^2 - 8x + 16 + 2y^2 + 4y + 2 = 18 \) \( x^2 + 2y^2 - 8x + 4y + 18 = 18 \) \( x^2 + 2y^2 - 8x + 4y = 0 \) This is the required equation of the ellipse. The distance between the center and focus helps set the orientation and specific parameters of the ellipse.In simple words: We used the center, one focus, and a point the ellipse goes through to figure out its exact equation. First, we found the distance to the focus, then used the point to solve for the ellipse's size, and finally put all the numbers into the ellipse formula.

🎯 Exam Tip: When the center and focus share a coordinate, it immediately tells you the orientation of the major axis. Remember that \(b^2 = a^2 - (ae)^2\) is crucial for finding \(b^2\). Always check for non-physical solutions like negative \(b^2\).

Question 9. Find the equation of the ellipse with its centre at (3, 1), vertex at (3, -2), and eccentricity equal to \( \frac{1}{3} \).
Answer: Let the center of the ellipse be \(C = (3, 1)\) and a vertex be \(A = (3, -2)\). Since the x-coordinates of the center and the vertex are the same (3), the major axis of the ellipse is parallel to the y-axis. The distance from the center to a vertex is \(a\) (the semi-major axis). \( a = CA = \sqrt{(3-3)^2 + (1 - (-2))^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \). So, \(a = 3\). Given eccentricity \(e = \frac{1}{3}\). Now, find \(b^2\) using the relationship \( b^2 = a^2 (1 - e^2) \): \( b^2 = 3^2 \left(1 - \left(\frac{1}{3}\right)^2\right) \) \( b^2 = 9 \left(1 - \frac{1}{9}\right) \) \( b^2 = 9 \left(\frac{9-1}{9}\right) \) \( b^2 = 9 \left(\frac{8}{9}\right) \) \( b^2 = 8 \) The general equation of an ellipse with its center \((h, k)\) and major axis parallel to the y-axis is: \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \) Substitute \(h=3\), \(k=1\), \(a^2=3^2=9\), and \(b^2=8\): \( \frac{(x-3)^2}{8} + \frac{(y-1)^2}{9} = 1 \) To clear the denominators, multiply the entire equation by the LCM of 8 and 9 (which is 72): \( 9(x-3)^2 + 8(y-1)^2 = 72 \) Expand the squared terms: \( 9(x^2 - 6x + 9) + 8(y^2 - 2y + 1) = 72 \) \( 9x^2 - 54x + 81 + 8y^2 - 16y + 8 = 72 \) Rearrange the terms: \( 9x^2 + 8y^2 - 54x - 16y + 89 = 72 \) \( 9x^2 + 8y^2 - 54x - 16y + 17 = 0 \) This is the required equation of the ellipse. Knowing the center, a vertex, and the eccentricity helps fully define the ellipse's shape and position.In simple words: We used the ellipse's center, a point on its edge (vertex), and its flatness value to find its equation. Since the center and vertex share an x-coordinate, the ellipse is tall. We found its long and short halves and then put them into the correct ellipse formula.

🎯 Exam Tip: The coordinates of the center and a vertex are critical for determining the semi-major axis \(a\) and the orientation of the ellipse. Ensure you use the correct form of the ellipse equation based on whether the major axis is horizontal or vertical.

Question 10. Find the equation of the ellipse whose centre is at (0, 2) and major axis along the axis of y and whose minor axis is equal to the distance between the foci and whose latus rectum is 2.
Answer: Let the center of the ellipse be \((h, k) = (0, 2)\). The major axis is along the y-axis, so the equation will be of the form \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \). Given that the minor axis is equal to the distance between the foci: \( 2b = 2ae \)
\( \implies \) \( b = ae \) ... (Equation 2) Given that the latus rectum is 2: \( \frac{2b^2}{a} = 2 \)
\( \implies \) \( b^2 = a \) ... (Equation 3) Now we have a system of equations. Substitute \(b = ae\) from (2) into (3): \( (ae)^2 = a \) \( a^2e^2 = a \) Since \(a \ne 0\) (it's a length), we can divide by \(a\): \( ae^2 = 1 \) ... (Equation 4) Now we have \(ae = b\) and \(ae^2 = 1\). Divide Equation 4 by Equation 2: \( \frac{ae^2}{ae} = \frac{1}{b} \) \( e = \frac{1}{b} \) Substitute \(e = \frac{1}{b}\) back into Equation 2 (\(b = ae\)): \( b = a \left(\frac{1}{b}\right) \) \( b^2 = a \) This confirms Equation 3. Now we have \(b^2 = a\) and \(ae^2 = 1\). From \(e = \frac{1}{b}\), we get \(e^2 = \frac{1}{b^2}\). We know \( b^2 = a \). So, \( e^2 = \frac{1}{a} \). Substitute this into \(ae^2 = 1\): \( a \left(\frac{1}{a}\right) = 1 \) This equation \(1=1\) doesn't help find \(a\) directly, it just shows consistency. Let's use the relation \( b^2 = a^2 (1 - e^2) \). Substitute \(b^2 = a\) and \(e^2 = \frac{1}{a}\) (since \(b^2=a\), \(e=\frac{1}{b}=\frac{1}{\sqrt{a}}\), so \(e^2=\frac{1}{a}\)): \( a = a^2 \left(1 - \frac{1}{a}\right) \) \( a = a^2 \left(\frac{a-1}{a}\right) \) \( a = a(a-1) \) Since \(a \ne 0\), we can divide by \(a\): \( 1 = a - 1 \)
\( \implies \) \( a = 2 \) Now find \(b^2\): \( b^2 = a = 2 \) So, \(a = 2\) and \(b^2 = 2\). Substitute \(h=0\), \(k=2\), \(a^2=2^2=4\), and \(b^2=2\) into the ellipse equation (\( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \)): \( \frac{(x-0)^2}{2} + \frac{(y-2)^2}{4} = 1 \) \( \frac{x^2}{2} + \frac{(y-2)^2}{4} = 1 \) To clear the denominators, multiply by the LCM of 2 and 4 (which is 4): \( 2x^2 + (y-2)^2 = 4 \) Expand and simplify: \( 2x^2 + (y^2 - 4y + 4) = 4 \) \( 2x^2 + y^2 - 4y + 4 = 4 \) \( 2x^2 + y^2 - 4y = 0 \) This is the required equation of the ellipse. Combining multiple given properties of an ellipse helps solve for its unknown parameters.In simple words: We found the equation for an ellipse that is centered at (0, 2) and tall (major axis along y-axis). We used the facts that its short side is equal to the distance between its foci and its latus rectum is 2. This helped us find the lengths of its major and minor axes, then write its full equation.

🎯 Exam Tip: When given multiple conditions, write them all down as equations involving \(a, b, e\). Then use algebraic substitution to solve for the unknown parameters. Be mindful of the orientation of the major axis when constructing the final equation.

Question 11. Find the equation of the ellipse with (i) focus at (1, -1), directrix x = 0, and e = \( \frac{\sqrt{2}}{2} \); (ii) focus at (0, 0), eccentricity is \( \frac{5}{6} \), and directrix is 3x + 4y - 1 = 0.
Answer: The definition of an ellipse states that for any point P(x, y) on the ellipse, the ratio of its distance from the focus (PF) to its distance from the directrix (PM) is constant and equal to the eccentricity (e). So, \( PF = e PM \). (i) Given focus \(F = (1, -1)\), directrix \(x = 0\), and eccentricity \(e = \frac{\sqrt{2}}{2}\). Let P(x, y) be any point on the ellipse. The distance PF is: \( \sqrt{(x-1)^2 + (y-(-1))^2} = \sqrt{(x-1)^2 + (y+1)^2} \). The distance PM from P(x, y) to the directrix \(x=0\) is \( |x| \). Apply the ellipse definition: \( \sqrt{(x-1)^2 + (y+1)^2} = \frac{\sqrt{2}}{2} |x| \). Square both sides of the equation to eliminate the square root: \( (x-1)^2 + (y+1)^2 = \left(\frac{\sqrt{2}}{2}\right)^2 x^2 \) \( (x-1)^2 + (y+1)^2 = \frac{2}{4} x^2 \) \( (x-1)^2 + (y+1)^2 = \frac{1}{2} x^2 \) Multiply by 2 to clear the fraction: \( 2[(x-1)^2 + (y+1)^2] = x^2 \) Expand the squared terms: \( 2(x^2 - 2x + 1 + y^2 + 2y + 1) = x^2 \) \( 2(x^2 + y^2 - 2x + 2y + 2) = x^2 \) \( 2x^2 + 2y^2 - 4x + 4y + 4 = x^2 \) Rearrange the terms to get the general equation of the ellipse: \( x^2 + 2y^2 - 4x + 4y + 4 = 0 \) This is the required equation of the ellipse. This method of using the focus-directrix property is a direct way to find the conic section's equation.(ii) Given focus \(F = (0, 0)\), eccentricity \(e = \frac{5}{6}\), and directrix \(3x + 4y - 1 = 0\). Let P(x, y) be any point on the ellipse. The distance PF is: \( \sqrt{(x-0)^2 + (y-0)^2} = \sqrt{x^2 + y^2} \). The distance PM from P(x, y) to the directrix \(Ax + By + C = 0\) is \( \frac{|Ax+By+C|}{\sqrt{A^2+B^2}} \). So, \( PM = \frac{|3x+4y-1|}{\sqrt{3^2+4^2}} = \frac{|3x+4y-1|}{\sqrt{9+16}} = \frac{|3x+4y-1|}{5} \). Apply the ellipse definition \( PF = e PM \): \( \sqrt{x^2+y^2} = \frac{5}{6} \left(\frac{|3x+4y-1|}{5}\right) \) \( \sqrt{x^2+y^2} = \frac{1}{6} |3x+4y-1| \) Square both sides: \( x^2+y^2 = \left(\frac{1}{6}\right)^2 (3x+4y-1)^2 \) \( x^2+y^2 = \frac{1}{36} (3x+4y-1)^2 \) Multiply by 36: \( 36(x^2+y^2) = (3x+4y-1)^2 \) Expand the right side (using \((A+B+C)^2 = A^2+B^2+C^2+2AB+2BC+2CA\)): \( 36x^2 + 36y^2 = (3x)^2 + (4y)^2 + (-1)^2 + 2(3x)(4y) + 2(4y)(-1) + 2(3x)(-1) \) \( 36x^2 + 36y^2 = 9x^2 + 16y^2 + 1 + 24xy - 8y - 6x \) Rearrange the terms to get the general equation of the ellipse: \( 36x^2 - 9x^2 + 36y^2 - 16y^2 - 24xy + 6x + 8y - 1 = 0 \) \( 27x^2 + 20y^2 - 24xy + 6x + 8y - 1 = 0 \) This is the required equation of the ellipse. These calculations allow for a direct conversion from geometric properties to an algebraic equation.In simple words: For both parts, we used the main rule of an ellipse: the distance from any point on the ellipse to a special point (focus) is always a fixed multiple (eccentricity) of its distance to a special line (directrix). We wrote this rule as an equation and then simplified it to get the ellipse's full equation.

🎯 Exam Tip: The formula \(PF = ePM\) is fundamental for these types of questions. Remember how to calculate the distance from a point to a line correctly, and be very careful with algebraic expansions and simplifications, especially when squaring trinomials.

Question 12. Find the equation of the ellipse from the following data : axis is coincident with x = 1, centre is (1, 5), focus is (1, 8) and the sum of the focal distances of a point on the ellipse is 12.
Answer: Let the center of the ellipse be \(C = (1, 5)\) and one focus be \(S = (1, 8)\). Since the x-coordinates of the center and the focus are the same (1), the major axis of the ellipse is parallel to the y-axis (vertical). The distance from the center to a focus is \(ae\): \( ae = CS = \sqrt{(1-1)^2 + (8-5)^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \). So, \(ae = 3\). The sum of the focal distances of any point on an ellipse is equal to \(2a\) (the length of the major axis). Given that the sum of focal distances is 12: \( 2a = 12 \)
\( \implies \) \( a = 6 \) Now we have \(a = 6\) and \(ae = 3\). We can find the eccentricity \(e\): \( (6)e = 3 \)
\( \implies \) \( e = \frac{3}{6} = \frac{1}{2} \) Next, find \(b^2\) using the relationship \( b^2 = a^2 (1 - e^2) \): \( b^2 = 6^2 \left(1 - \left(\frac{1}{2}\right)^2\right) \) \( b^2 = 36 \left(1 - \frac{1}{4}\right) \) \( b^2 = 36 \left(\frac{3}{4}\right) \) \( b^2 = 9 \times 3 \) \( b^2 = 27 \) The general equation of an ellipse with its center \((h, k)\) and major axis parallel to the y-axis is: \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \) Substitute \(h=1\), \(k=5\), \(a^2=6^2=36\), and \(b^2=27\): \( \frac{(x-1)^2}{27} + \frac{(y-5)^2}{36} = 1 \) This is the required equation of the ellipse. The sum of focal distances provides a direct way to find the major axis length.In simple words: We found the equation for an ellipse. We knew its center, one focus, and that the sum of distances from any point on it to the two foci is always 12. Since the center and focus had the same x-coordinate, we knew the ellipse was tall. We used these facts to find its size and then wrote down its equation.

🎯 Exam Tip: The properties \(CS=ae\) and \(PS+PS'=2a\) are fundamental for solving these problems. Correctly identifying the major axis orientation (vertical in this case) is crucial for setting up the equation.

Question 13. A point P (x, y) moves so that the product of the slopes of the two lines joining P to the two points (-2, 1) and (6, 5) is -4. Show that the locus is an ellipse and locate its centre.
Answer: Let the moving point be \(P(x, y)\). The two fixed points are \(A(-2, 1)\) and \(B(6, 5)\). The slope of the line joining \(P(x, y)\) to \(A(-2, 1)\) is \( m_{PA} = \frac{y-1}{x-(-2)} = \frac{y-1}{x+2} \). The slope of the line joining \(P(x, y)\) to \(B(6, 5)\) is \( m_{PB} = \frac{y-5}{x-6} \). According to the given condition, the product of these slopes is -4: \( m_{PA} \times m_{PB} = -4 \) \( \left(\frac{y-1}{x+2}\right) \left(\frac{y-5}{x-6}\right) = -4 \) Multiply both sides by \((x+2)(x-6)\): \( (y-1)(y-5) = -4(x+2)(x-6) \) Expand both sides: \( y^2 - 5y - y + 5 = -4(x^2 - 6x + 2x - 12) \) \( y^2 - 6y + 5 = -4(x^2 - 4x - 12) \) \( y^2 - 6y + 5 = -4x^2 + 16x + 48 \) Rearrange the terms to bring all terms to one side: \( 4x^2 + y^2 - 16x - 6y + 5 - 48 = 0 \) \( 4x^2 + y^2 - 16x - 6y - 43 = 0 \) This is the general equation of a conic section. To show it's an ellipse and find its center, we complete the square for both \(x\) and \(y\) terms: \( 4(x^2 - 4x) + (y^2 - 6y) = 43 \) Complete the square for \(x\) by adding and subtracting \( (-4/2)^2 = (-2)^2 = 4 \): Complete the square for \(y\) by adding and subtracting \( (-6/2)^2 = (-3)^2 = 9 \): \( 4(x^2 - 4x + 4 - 4) + (y^2 - 6y + 9 - 9) = 43 \) \( 4((x-2)^2 - 4) + ((y-3)^2 - 9) = 43 \) \( 4(x-2)^2 - 16 + (y-3)^2 - 9 = 43 \) Move the constant terms to the right side: \( 4(x-2)^2 + (y-3)^2 = 43 + 16 + 9 \) \( 4(x-2)^2 + (y-3)^2 = 68 \) To get the standard form of an ellipse, divide the entire equation by 68: \( \frac{4(x-2)^2}{68} + \frac{(y-3)^2}{68} = 1 \) \( \frac{(x-2)^2}{17} + \frac{(y-3)^2}{68} = 1 \) This equation is in the standard form \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \) (since \(68 > 17\), so \(a^2=68\) and \(b^2=17\)), which represents an ellipse. Comparing with the standard form, the center of the ellipse is \((h, k) = (2, 3)\). The locus of point P is an ellipse with its center at \((2, 3)\).In simple words: We used the condition about the slopes of lines from point P to two other points. This led us to a complex equation. By rearranging and using a math trick called "completing the square," we changed it into the standard form of an ellipse, which showed us that the point P traces an ellipse with its center at (2, 3).

🎯 Exam Tip: This question tests your ability to translate a geometric condition into an algebraic equation and then manipulate it into the standard form of a conic section. Completing the square is a critical skill for identifying the type of conic and its center.

Question 14. Find the eccentricity, the coordinates of the foci, and the length of the latus rectum of the ellipse \( 2x^2 + 3y^2 = 1 \).
Answer: The given equation of the ellipse is \( 2x^2 + 3y^2 = 1 \). To find its properties, we first write it in standard form: Divide by 1: \( \frac{2x^2}{1} + \frac{3y^2}{1} = 1 \) This gives: \( \frac{x^2}{1/2} + \frac{y^2}{1/3} = 1 \) By comparing this with the standard equation of an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (where \( a > b \)), we see that \( a^2 = \frac{1}{2} \) and \( b^2 = \frac{1}{3} \). Since \( a^2 = \frac{1}{2} \), we have \( a = \frac{1}{\sqrt{2}} \). We know the relation \( b^2 = a^2(1 - e^2) \) where \( e \) is the eccentricity. Substitute the values: \( \frac{1}{3} = \frac{1}{2}(1 - e^2) \) Multiply both sides by 2: \( \frac{2}{3} = 1 - e^2 \) Rearrange to find \( e^2 \): \( e^2 = 1 - \frac{2}{3} = \frac{1}{3} \) So, the eccentricity \( e = \frac{1}{\sqrt{3}} \) (since \( e > 0 \)). The foci of the ellipse are given by \( (\pm ae, 0) \). Substitute \( a = \frac{1}{\sqrt{2}} \) and \( e = \frac{1}{\sqrt{3}} \): \( ae = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{6}} \) So, the coordinates of the foci are \( \left(\pm \frac{1}{\sqrt{6}}, 0\right) \). The length of the latus rectum is given by \( \frac{2b^2}{a} \). Substitute \( b^2 = \frac{1}{3} \) and \( a = \frac{1}{\sqrt{2}} \): Length of latus rectum \( = \frac{2 \times \frac{1}{3}}{\frac{1}{\sqrt{2}}} = \frac{2}{3} \times \sqrt{2} = \frac{2\sqrt{2}}{3} \).In simple words: First, we change the ellipse equation to a standard form. Then, we find 'a' and 'b' which are like the half-lengths of the ellipse. After that, we calculate the eccentricity 'e', which tells us how flat the ellipse is. Finally, we use 'a', 'b', and 'e' to find where the two special points (foci) are and the length of a specific chord (latus rectum) across the ellipse.

🎯 Exam Tip: Remember to always convert the given equation to the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) or \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) first, to correctly identify 'a' and 'b' before calculating other properties like eccentricity, foci, and latus rectum.

Question 15. For the ellipse, \( 9x^2 + 16y^2 = 576 \), find the semi-major axis, the semi-minor axis, the eccentricity, the coordinates of the foci, the equations of the directrices, and the length of the latus rectum.
Answer: The given equation of the ellipse is \( 9x^2 + 16y^2 = 576 \). To find its properties, we first write it in standard form: Divide by 576: \( \frac{9x^2}{576} + \frac{16y^2}{576} = 1 \) This simplifies to: \( \frac{x^2}{64} + \frac{y^2}{36} = 1 \) By comparing this with the standard equation of an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (where \( a > b \)), we have: \( a^2 = 64 \implies a = 8 \) (This is the semi-major axis). \( b^2 = 36 \implies b = 6 \) (This is the semi-minor axis). Now we find the eccentricity \( e \) using the relation \( b^2 = a^2(1 - e^2) \): \( 36 = 64(1 - e^2) \) Divide by 64: \( \frac{36}{64} = 1 - e^2 \) Simplify the fraction: \( \frac{9}{16} = 1 - e^2 \) Rearrange to find \( e^2 \): \( e^2 = 1 - \frac{9}{16} = \frac{16 - 9}{16} = \frac{7}{16} \) So, the eccentricity \( e = \frac{\sqrt{7}}{4} \) (since \( e > 0 \)). The coordinates of the foci are given by \( (\pm ae, 0) \): Substitute \( a = 8 \) and \( e = \frac{\sqrt{7}}{4} \): \( ae = 8 \times \frac{\sqrt{7}}{4} = 2\sqrt{7} \) So, the foci are at \( (\pm 2\sqrt{7}, 0) \). The equations of the directrices are given by \( x = \pm \frac{a}{e} \): Substitute \( a = 8 \) and \( e = \frac{\sqrt{7}}{4} \): \( \frac{a}{e} = \frac{8}{\frac{\sqrt{7}}{4}} = 8 \times \frac{4}{\sqrt{7}} = \frac{32}{\sqrt{7}} \) So, the directrices are \( x = \pm \frac{32}{\sqrt{7}} \). The length of the latus rectum is given by \( \frac{2b^2}{a} \): Substitute \( b^2 = 36 \) and \( a = 8 \): Length of latus rectum \( = \frac{2 \times 36}{8} = \frac{72}{8} = 9 \).In simple words: We first rewrite the ellipse equation to find its half-lengths (semi-major and semi-minor axes). Then, we calculate its eccentricity, which shows how elongated it is. Using these values, we can locate the two special points (foci) and define the guide lines (directrices). Lastly, we find the length of a specific line segment called the latus rectum.

🎯 Exam Tip: When finding properties of an ellipse, always start by normalizing the equation to its standard form. Pay close attention to whether \( a^2 \) is under the \( x^2 \) or \( y^2 \) term, as this determines if the major axis is horizontal or vertical, affecting the coordinates of the foci and directrices.

Question 16. Find the length of the axes, the co-ordinates of the foci, the eccentricity, and latus rectum of the ellipse \( 3x^2 + 2y^2 = 24 \).
Answer: The given equation of the ellipse is \( 3x^2 + 2y^2 = 24 \). To find its properties, we first write it in standard form: Divide by 24: \( \frac{3x^2}{24} + \frac{2y^2}{24} = 1 \) This simplifies to: \( \frac{x^2}{8} + \frac{y^2}{12} = 1 \) By comparing this with the standard equation of an ellipse \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) (since the larger denominator is under \( y^2 \), indicating a vertical major axis), we have: \( a^2 = 12 \implies a = \sqrt{12} = 2\sqrt{3} \) (This is the semi-major axis). \( b^2 = 8 \implies b = \sqrt{8} = 2\sqrt{2} \) (This is the semi-minor axis). The length of the major axis is \( 2a = 2 \times 2\sqrt{3} = 4\sqrt{3} \). The length of the minor axis is \( 2b = 2 \times 2\sqrt{2} = 4\sqrt{2} \). Now we find the eccentricity \( e \) using the relation \( b^2 = a^2(1 - e^2) \): \( 8 = 12(1 - e^2) \) Divide by 12: \( \frac{8}{12} = 1 - e^2 \) Simplify the fraction: \( \frac{2}{3} = 1 - e^2 \) Rearrange to find \( e^2 \): \( e^2 = 1 - \frac{2}{3} = \frac{1}{3} \) So, the eccentricity \( e = \frac{1}{\sqrt{3}} \) (since \( e > 0 \)). The coordinates of the foci are given by \( (0, \pm ae) \) for a vertical major axis: Substitute \( a = 2\sqrt{3} \) and \( e = \frac{1}{\sqrt{3}} \): \( ae = 2\sqrt{3} \times \frac{1}{\sqrt{3}} = 2 \) So, the foci are at \( (0, \pm 2) \). The length of the latus rectum is given by \( \frac{2b^2}{a} \): Substitute \( b^2 = 8 \) and \( a = 2\sqrt{3} \): Length of latus rectum \( = \frac{2 \times 8}{2\sqrt{3}} = \frac{16}{2\sqrt{3}} = \frac{8}{\sqrt{3}} \).In simple words: First, we convert the ellipse equation to a standard form. Then, we find the full lengths of the ellipse's widest and narrowest parts (major and minor axes). Next, we calculate how oval-shaped the ellipse is (eccentricity) and use this to find the two special points inside it (foci). Finally, we determine the length of the latus rectum, which is a specific chord through a focus.

🎯 Exam Tip: For ellipses, always correctly identify which axis is major (longer) and which is minor (shorter) by comparing the denominators in the standard form. If \( a^2 \) is under \( y^2 \), the major axis is vertical, and foci will be on the y-axis, not the x-axis.

Question 17. Find the eccentricity of the ellipse, \( 4x^2 + 9y^2 - 8x - 36y + 4 = 0 \).
Answer: The given equation of the ellipse is \( 4x^2 + 9y^2 - 8x - 36y + 4 = 0 \). To find its eccentricity, we first need to transform the equation into standard form by completing the square: Group x-terms and y-terms: \( (4x^2 - 8x) + (9y^2 - 36y) = -4 \) Factor out coefficients of \( x^2 \) and \( y^2 \): \( 4(x^2 - 2x) + 9(y^2 - 4y) = -4 \) Complete the square for x-terms (\( (x-1)^2 = x^2 - 2x + 1 \)): \( 4(x^2 - 2x + 1 - 1) + 9(y^2 - 4y) = -4 \) Complete the square for y-terms (\( (y-2)^2 = y^2 - 4y + 4 \)): \( 4(x^2 - 2x + 1 - 1) + 9(y^2 - 4y + 4 - 4) = -4 \) Rewrite using squared terms: \( 4[(x - 1)^2 - 1] + 9[(y - 2)^2 - 4] = -4 \) Distribute coefficients: \( 4(x - 1)^2 - 4 + 9(y - 2)^2 - 36 = -4 \) Combine constant terms: \( 4(x - 1)^2 + 9(y - 2)^2 - 40 = -4 \) Move the constant to the right side: \( 4(x - 1)^2 + 9(y - 2)^2 = -4 + 40 \) \( 4(x - 1)^2 + 9(y - 2)^2 = 36 \) Now, divide by 36 to get the standard form: \( \frac{4(x - 1)^2}{36} + \frac{9(y - 2)^2}{36} = 1 \) This simplifies to: \( \frac{(x - 1)^2}{9} + \frac{(y - 2)^2}{4} = 1 \) By comparing this with the standard equation of an ellipse \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) (where \( a > b \)), we have: \( a^2 = 9 \implies a = 3 \) \( b^2 = 4 \implies b = 2 \) Now we find the eccentricity \( e \) using the relation \( b^2 = a^2(1 - e^2) \): \( 4 = 9(1 - e^2) \) Divide by 9: \( \frac{4}{9} = 1 - e^2 \) Rearrange to find \( e^2 \): \( e^2 = 1 - \frac{4}{9} = \frac{9 - 4}{9} = \frac{5}{9} \) So, the eccentricity \( e = \frac{\sqrt{5}}{3} \) (since \( e > 0 \)).In simple words: We start with the given equation and rearrange it by completing the square for both the x and y terms. This transforms the messy equation into a clear, standard form of an ellipse. From this standard form, we can easily find 'a' and 'b', which are the semi-axes. Finally, we use a special formula that connects 'a', 'b', and 'e' to find the eccentricity 'e', which tells us how round or stretched the ellipse is.

🎯 Exam Tip: Completing the square is a key technique for general conic section equations. Be careful with signs and distributing coefficients when moving terms around to get the standard form. The eccentricity will always be between 0 and 1 for an ellipse.

Question 18. Find the centre of the ellipse, \( \frac{x^2-ax}{a^2} + \frac{y^2-by}{b^2} = 0 \).
Answer: The given equation of the ellipse is \( \frac{x^2-ax}{a^2} + \frac{y^2-by}{b^2} = 0 \). To find the center, we need to rewrite this equation in the standard form \( \frac{(x-h)^2}{A^2} + \frac{(y-k)^2}{B^2} = 1 \), where \( (h, k) \) is the center. First, multiply the entire equation by \( a^2 b^2 \) to clear denominators: \( b^2(x^2 - ax) + a^2(y^2 - by) = 0 \) Now, we complete the square for the x-terms and y-terms: For x-terms: \( x^2 - ax \). To complete the square, we add and subtract \( \left(\frac{-a}{2}\right)^2 = \frac{a^2}{4} \). For y-terms: \( y^2 - by \). To complete the square, we add and subtract \( \left(\frac{-b}{2}\right)^2 = \frac{b^2}{4} \). Substitute these into the equation: \( b^2\left(x^2 - ax + \frac{a^2}{4} - \frac{a^2}{4}\right) + a^2\left(y^2 - by + \frac{b^2}{4} - \frac{b^2}{4}\right) = 0 \) Group the perfect square trinomials: \( b^2\left[\left(x - \frac{a}{2}\right)^2 - \frac{a^2}{4}\right] + a^2\left[\left(y - \frac{b}{2}\right)^2 - \frac{b^2}{4}\right] = 0 \) Distribute \( b^2 \) and \( a^2 \): \( b^2\left(x - \frac{a}{2}\right)^2 - \frac{b^2 a^2}{4} + a^2\left(y - \frac{b}{2}\right)^2 - \frac{a^2 b^2}{4} = 0 \) Combine constant terms and move them to the right side: \( b^2\left(x - \frac{a}{2}\right)^2 + a^2\left(y - \frac{b}{2}\right)^2 = \frac{b^2 a^2}{4} + \frac{a^2 b^2}{4} \) \( b^2\left(x - \frac{a}{2}\right)^2 + a^2\left(y - \frac{b}{2}\right)^2 = \frac{2a^2 b^2}{4} \) \( b^2\left(x - \frac{a}{2}\right)^2 + a^2\left(y - \frac{b}{2}\right)^2 = \frac{a^2 b^2}{2} \) Now, divide the entire equation by \( \frac{a^2 b^2}{2} \) to get the standard form: \( \frac{b^2\left(x - \frac{a}{2}\right)^2}{\frac{a^2 b^2}{2}} + \frac{a^2\left(y - \frac{b}{2}\right)^2}{\frac{a^2 b^2}{2}} = 1 \) This simplifies to: \( \frac{\left(x - \frac{a}{2}\right)^2}{\frac{a^2}{2}} + \frac{\left(y - \frac{b}{2}\right)^2}{\frac{b^2}{2}} = 1 \) Comparing this with \( \frac{(x-h)^2}{A^2} + \frac{(y-k)^2}{B^2} = 1 \), the center \( (h, k) \) is \( \left(\frac{a}{2}, \frac{b}{2}\right) \).In simple words: To find the center of this ellipse, we first get rid of the fractions in the equation. Then, we use a method called "completing the square" for both the x and y parts of the equation. This helps us rewrite the equation in a special form that directly shows the coordinates of the ellipse's center. The center ends up being at \( \left(\frac{a}{2}, \frac{b}{2}\right) \).

🎯 Exam Tip: When an ellipse equation is not in standard form, completing the square is essential. Remember that the center of the ellipse is found from the \( (x-h)^2 \) and \( (y-k)^2 \) terms, so careful algebraic manipulation is key.

Question 19. Find the distance between a focus and an extremity of the minor axis of the ellipse
(i) \( 4x^2 + 5y^2 = 100 \)
(ii) \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
Answer:
(i) Given equation of the ellipse: \( 4x^2 + 5y^2 = 100 \). First, convert to standard form by dividing by 100: \( \frac{4x^2}{100} + \frac{5y^2}{100} = 1 \) \( \frac{x^2}{25} + \frac{y^2}{20} = 1 \) Comparing with \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we have \( a^2 = 25 \) and \( b^2 = 20 \). So, \( a = 5 \). To find the eccentricity \( e \), use \( b^2 = a^2(1 - e^2) \): \( 20 = 25(1 - e^2) \) \( \frac{20}{25} = 1 - e^2 \) \( \frac{4}{5} = 1 - e^2 \) \( e^2 = 1 - \frac{4}{5} = \frac{1}{5} \) So, \( e = \frac{1}{\sqrt{5}} \). The foci are at \( (\pm ae, 0) \). \( ae = 5 \times \frac{1}{\sqrt{5}} = \sqrt{5} \). Let's consider the focus at \( (\sqrt{5}, 0) \). The extremities of the minor axis are \( (0, \pm b) \). \( b = \sqrt{20} = 2\sqrt{5} \). Let's consider the extremity of the minor axis at \( (0, 2\sqrt{5}) \). The distance between a focus \( (\sqrt{5}, 0) \) and an extremity of the minor axis \( (0, 2\sqrt{5}) \) is calculated using the distance formula: Distance \( = \sqrt{(\sqrt{5} - 0)^2 + (0 - 2\sqrt{5})^2} \) \( = \sqrt{(\sqrt{5})^2 + (-2\sqrt{5})^2} \) \( = \sqrt{5 + (4 \times 5)} = \sqrt{5 + 20} = \sqrt{25} = 5 \). The distance between a focus and an extremity of the minor axis is always equal to the semi-major axis, which is 'a'. Here, \( a=5 \). (ii) Given standard equation of the ellipse: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). The foci are at \( (\pm ae, 0) \). Let's take one focus as \( F(ae, 0) \). The extremities of the minor axis are \( (0, \pm b) \). Let's take one extremity as \( B(0, b) \). The distance between the focus \( F(ae, 0) \) and the extremity \( B(0, b) \) is: Distance \( = \sqrt{(ae - 0)^2 + (0 - b)^2} \) \( = \sqrt{(ae)^2 + (-b)^2} \) \( = \sqrt{a^2e^2 + b^2} \) We know the relation for an ellipse: \( b^2 = a^2(1 - e^2) \), which means \( b^2 = a^2 - a^2e^2 \), or \( a^2e^2 = a^2 - b^2 \). Substitute \( a^2e^2 = a^2 - b^2 \) into the distance formula: Distance \( = \sqrt{(a^2 - b^2) + b^2} = \sqrt{a^2} = a \). So, the distance between a focus and an extremity of the minor axis is always equal to the length of the semi-major axis, \( a \).In simple words: For any ellipse, the distance from one of its special points (a focus) to the very end of its shorter side (an extremity of the minor axis) is always the same as the length of its semi-major axis, which is 'a'. We calculate 'a', 'b', and 'e' from the given equation, find the coordinates of a focus and a minor axis extremity, then use the distance formula. For the general form, we proved that this distance simplifies directly to 'a'.

🎯 Exam Tip: Remember the property that the distance from a focus to an extremity of the minor axis is equal to the semi-major axis (a). This can save time in calculations, but you should also know how to derive it using the distance formula and the relationship \( b^2 = a^2(1 - e^2) \).

Question 20. Given the ellipse \( 36x^2 + 100y^2 = 3600 \), find the equation and the lengths of the focal radii drawn through the point \( \left(8, \frac{18}{5}\right) \).
Answer: The given equation of the ellipse is \( 36x^2 + 100y^2 = 3600 \). First, convert to standard form by dividing by 3600: \( \frac{36x^2}{3600} + \frac{100y^2}{3600} = 1 \) \( \frac{x^2}{100} + \frac{y^2}{36} = 1 \) Comparing with \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we have \( a^2 = 100 \) and \( b^2 = 36 \). So, \( a = 10 \) and \( b = 6 \). To find the eccentricity \( e \), use \( b^2 = a^2(1 - e^2) \): \( 36 = 100(1 - e^2) \) \( \frac{36}{100} = 1 - e^2 \) \( \frac{9}{25} = 1 - e^2 \) \( e^2 = 1 - \frac{9}{25} = \frac{16}{25} \) So, \( e = \frac{4}{5} \). The foci are at \( (\pm ae, 0) \). \( ae = 10 \times \frac{4}{5} = 8 \). So, the foci are \( F_1(8, 0) \) and \( F_2(-8, 0) \). Let the given point be \( P\left(8, \frac{18}{5}\right) \). The focal radii are the distances from the point P to each focus. Length of focal radius \( PF_1 \) (distance from \( P\left(8, \frac{18}{5}\right) \) to \( F_1(8, 0) \)): \( PF_1 = \sqrt{(8-8)^2 + \left(\frac{18}{5} - 0\right)^2} = \sqrt{0^2 + \left(\frac{18}{5}\right)^2} = \frac{18}{5} \). Length of focal radius \( PF_2 \) (distance from \( P\left(8, \frac{18}{5}\right) \) to \( F_2(-8, 0) \)): \( PF_2 = \sqrt{(8 - (-8))^2 + \left(\frac{18}{5} - 0\right)^2} \) \( = \sqrt{(16)^2 + \left(\frac{18}{5}\right)^2} = \sqrt{256 + \frac{324}{25}} \) \( = \sqrt{\frac{256 \times 25 + 324}{25}} = \sqrt{\frac{6400 + 324}{25}} = \sqrt{\frac{6724}{25}} = \frac{82}{5} \). Alternatively, the lengths of focal radii are \( a \pm ex \). \( PF_1 = a - ex = 10 - \frac{4}{5}(8) = 10 - \frac{32}{5} = \frac{50 - 32}{5} = \frac{18}{5} \). \( PF_2 = a + ex = 10 + \frac{4}{5}(8) = 10 + \frac{32}{5} = \frac{50 + 32}{5} = \frac{82}{5} \). Now, we find the equations of the focal radii (lines joining P to each focus). Equation of line \( PF_1 \) (passing through \( P\left(8, \frac{18}{5}\right) \) and \( F_1(8, 0) \)): This is a vertical line since the x-coordinates are the same. The equation is \( x = 8 \). Equation of line \( PF_2 \) (passing through \( P\left(8, \frac{18}{5}\right) \) and \( F_2(-8, 0) \)): Using the two-point form \( y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1) \): \( y - 0 = \frac{\frac{18}{5} - 0}{8 - (-8)}(x - (-8)) \) \( y = \frac{\frac{18}{5}}{16}(x + 8) \) \( y = \frac{18}{5 \times 16}(x + 8) \) \( y = \frac{9}{40}(x + 8) \) Multiply by 40: \( 40y = 9(x + 8) \) \( 40y = 9x + 72 \) Rearrange: \( 9x - 40y + 72 = 0 \).In simple words: First, we convert the ellipse equation to its standard form to find its half-lengths and how stretched it is (eccentricity). This helps us locate the two special points called foci. Then, for the given point, we find the distance to each focus (focal radii lengths) using a formula. Finally, we find the equations of the straight lines that connect the given point to each of the foci.

🎯 Exam Tip: Remember that focal radii lengths for a point \( (x, y) \) on an ellipse with major axis along x-axis are \( a \pm ex \). One focal radius is \( a - ex \) and the other is \( a + ex \). For equations of lines, use the two-point form of a straight line if needed.

Question 21. The focal distance of an end of the minor axis of the ellipse is k and the distance between the foci is 2h. Find the lengths of the semi-axes.
Answer: Let \( a \) be the semi-major axis and \( b \) be the semi-minor axis of the ellipse, and \( e \) be its eccentricity. The ends of the minor axis are \( (0, \pm b) \). The foci are \( (\pm ae, 0) \). The focal distance of an end of the minor axis is the distance from \( (0, b) \) to a focus, say \( (ae, 0) \). We know that this distance is equal to the semi-major axis, \( a \). So, according to the given condition, the focal distance is \( k \). Therefore, \( a = k \). This is the length of the semi-major axis. Next, the distance between the foci is given as \( 2h \). We know that the distance between the foci is \( 2ae \). So, \( 2ae = 2h \), which means \( ae = h \). Now we have \( a = k \) and \( ae = h \). Substitute \( a = k \) into the second equation: \( k \times e = h \implies e = \frac{h}{k} \). We also know the relationship between \( a, b, \) and \( e \) for an ellipse: \( b^2 = a^2(1 - e^2) \). Substitute \( a = k \) and \( e = \frac{h}{k} \): \( b^2 = k^2\left(1 - \left(\frac{h}{k}\right)^2\right) \) \( b^2 = k^2\left(1 - \frac{h^2}{k^2}\right) \) \( b^2 = k^2\left(\frac{k^2 - h^2}{k^2}\right) \) \( b^2 = k^2 - h^2 \) So, the length of the semi-minor axis \( b = \sqrt{k^2 - h^2} \). Thus, the lengths of the semi-axes are: Semi-major axis \( a = k \) Semi-minor axis \( b = \sqrt{k^2 - h^2} \)In simple words: We are given two pieces of information about an ellipse: the distance from the end of its shorter side to a special point (focus), and the distance between the two special points (foci). We know that the first distance is always equal to the semi-major axis, so that gives us 'a'. The second distance gives us 'ae'. Using these, we can find the eccentricity 'e'. Finally, we use the main formula that relates 'a', 'b', and 'e' to find the length of the semi-minor axis 'b' in terms of 'k' and 'h'.

🎯 Exam Tip: Remember the fundamental properties of an ellipse: the focal distance from an end of the minor axis is 'a', and the distance between foci is '2ae'. Using these two facts is the most straightforward way to solve such problems.

Question 22. Find the eccentricity of the ellipse whose latus rectum is 4 and distance of the vertex from the nearest focus is 1.5 cm.
Answer: Let \( a \) be the semi-major axis, \( b \) be the semi-minor axis, and \( e \) be the eccentricity of the ellipse. Given that the length of the latus rectum is 4. The formula for the latus rectum is \( \frac{2b^2}{a} \). So, \( \frac{2b^2}{a} = 4 \). This implies \( 2b^2 = 4a \), which simplifies to \( b^2 = 2a \) ...(1) Given that the distance of the vertex from the nearest focus is 1.5 cm. For an ellipse with the major axis along the x-axis, the vertices are \( (\pm a, 0) \) and the foci are \( (\pm ae, 0) \). The nearest vertex to the focus \( (ae, 0) \) is \( (a, 0) \). The distance between \( (a, 0) \) and \( (ae, 0) \) is \( a - ae \). So, \( a - ae = 1.5 \). This can be written as \( a(1 - e) = 1.5 \) ...(2) We also know the fundamental relation for an ellipse: \( b^2 = a^2(1 - e^2) \). Substitute \( b^2 = 2a \) from (1): \( 2a = a^2(1 - e^2) \) Since \( a \) cannot be 0 (it's a length), we can divide both sides by \( a \): \( 2 = a(1 - e^2) \) We know that \( 1 - e^2 = (1 - e)(1 + e) \). So, \( 2 = a(1 - e)(1 + e) \) From equation (2), we know \( a(1 - e) = 1.5 \). Substitute this into the equation: \( 2 = 1.5(1 + e) \) Now, solve for \( e \): \( \frac{2}{1.5} = 1 + e \) \( \frac{2}{\frac{3}{2}} = 1 + e \) \( \frac{4}{3} = 1 + e \) \( e = \frac{4}{3} - 1 \) \( e = \frac{4 - 3}{3} = \frac{1}{3} \). Thus, the eccentricity of the ellipse is \( \frac{1}{3} \).In simple words: We are given the length of the latus rectum and the distance from a vertex to the closest focus. We use the formulas for both of these properties, which involve 'a' (semi-major axis), 'b' (semi-minor axis), and 'e' (eccentricity). By substituting these into the main ellipse equation that connects 'a', 'b', and 'e', we create a system of equations. Solving this system allows us to find the value of 'e', which tells us how oval-shaped the ellipse is.

🎯 Exam Tip: This problem requires combining multiple properties of the ellipse. Make sure you recall the formulas for latus rectum and the distance between a vertex and a focus correctly. Factorization of \( (1 - e^2) \) into \( (1 - e)(1 + e) \) is often a useful algebraic step in such problems.

Question 23. The directrix of a conic section is the line \( 3x + 4y = 1 \) and the focus S is \( (-2, 3) \). If the eccentricity e is \( \frac{1}{\sqrt{2}} \), find the equation to the conic section.
Answer: Let \( P(x, y) \) be any point on the conic section. The focus \( F \) is at \( (-2, 3) \). The equation of the directrix \( L \) is \( 3x + 4y - 1 = 0 \). The eccentricity \( e = \frac{1}{\sqrt{2}} \). By the definition of a conic section, the distance from any point \( P \) on the conic to the focus \( F \) is \( e \) times the perpendicular distance from \( P \) to the directrix \( L \). So, \( PF = e \times PM \). The distance \( PF = \sqrt{(x - (-2))^2 + (y - 3)^2} = \sqrt{(x + 2)^2 + (y - 3)^2} \). The perpendicular distance \( PM \) from \( P(x, y) \) to the line \( 3x + 4y - 1 = 0 \) is: \( PM = \frac{|3x + 4y - 1|}{\sqrt{3^2 + 4^2}} = \frac{|3x + 4y - 1|}{\sqrt{9 + 16}} = \frac{|3x + 4y - 1|}{\sqrt{25}} = \frac{|3x + 4y - 1|}{5} \). Now, substitute these into \( PF = e \times PM \): \( \sqrt{(x + 2)^2 + (y - 3)^2} = \frac{1}{\sqrt{2}} \times \frac{|3x + 4y - 1|}{5} \) To remove the square root, square both sides of the equation: \( (x + 2)^2 + (y - 3)^2 = \left(\frac{1}{\sqrt{2}}\right)^2 \times \left(\frac{3x + 4y - 1}{5}\right)^2 \) \( (x^2 + 4x + 4) + (y^2 - 6y + 9) = \frac{1}{2} \times \frac{(3x + 4y - 1)^2}{25} \) \( x^2 + y^2 + 4x - 6y + 13 = \frac{1}{50}(9x^2 + 16y^2 + 1 + 24xy - 6x - 8y) \) Multiply both sides by 50: \( 50(x^2 + y^2 + 4x - 6y + 13) = 9x^2 + 16y^2 + 1 + 24xy - 6x - 8y \) \( 50x^2 + 50y^2 + 200x - 300y + 650 = 9x^2 + 16y^2 + 24xy - 6x - 8y + 1 \) Move all terms to one side to get the general equation: \( (50x^2 - 9x^2) + (50y^2 - 16y^2) - 24xy + (200x + 6x) + (-300y + 8y) + (650 - 1) = 0 \) \( 41x^2 + 34y^2 - 24xy + 206x - 292y + 649 = 0 \). This is the required equation of the conic section. Since \( e = \frac{1}{\sqrt{2}} < 1 \), the conic section is an ellipse.In simple words: We find the equation of a conic by using its basic definition: any point on the conic is a certain distance from a special point (the focus) and a special line (the directrix). This distance relationship is controlled by the eccentricity 'e'. We write out the distance formulas, substitute the given focus, directrix, and eccentricity, then square both sides to remove square roots. Finally, we rearrange the terms to get the general equation of the conic. Since 'e' is less than 1, this conic is an ellipse.

🎯 Exam Tip: The definition \( PF = e \times PM \) is fundamental for finding the equation of any conic section when given a focus, directrix, and eccentricity. Be meticulous with squaring both sides and expanding polynomial terms, as algebraic errors are common here.

Question 24. Find the equation to the conic section whose focus is \( (1, -1) \), eccentricity is \( \frac{1}{2} \) and the directrix is the line \( x - y = 3 \). Is the conic section an ellipse ?
Answer: Let \( P(x, y) \) be any point on the conic section. The focus \( F \) is at \( (1, -1) \). The equation of the directrix \( L \) is \( x - y - 3 = 0 \). The eccentricity \( e = \frac{1}{2} \). By the definition of a conic section, the distance from any point \( P \) on the conic to the focus \( F \) is \( e \) times the perpendicular distance from \( P \) to the directrix \( L \). So, \( PF = e \times PM \). The distance \( PF = \sqrt{(x - 1)^2 + (y - (-1))^2} = \sqrt{(x - 1)^2 + (y + 1)^2} \). The perpendicular distance \( PM \) from \( P(x, y) \) to the line \( x - y - 3 = 0 \) is: \( PM = \frac{|x - y - 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y - 3|}{\sqrt{1 + 1}} = \frac{|x - y - 3|}{\sqrt{2}} \). Now, substitute these into \( PF = e \times PM \): \( \sqrt{(x - 1)^2 + (y + 1)^2} = \frac{1}{2} \times \frac{|x - y - 3|}{\sqrt{2}} \) To remove the square root, square both sides of the equation: \( (x - 1)^2 + (y + 1)^2 = \left(\frac{1}{2}\right)^2 \times \left(\frac{x - y - 3}{\sqrt{2}}\right)^2 \) \( (x^2 - 2x + 1) + (y^2 + 2y + 1) = \frac{1}{4} \times \frac{(x - y - 3)^2}{2} \) \( x^2 + y^2 - 2x + 2y + 2 = \frac{1}{8}(x^2 + y^2 + 9 - 2xy - 6x + 6y) \) Multiply both sides by 8: \( 8(x^2 + y^2 - 2x + 2y + 2) = x^2 + y^2 - 2xy - 6x + 6y + 9 \) \( 8x^2 + 8y^2 - 16x + 16y + 16 = x^2 + y^2 - 2xy - 6x + 6y + 9 \) Move all terms to one side to get the general equation: \( (8x^2 - x^2) + (8y^2 - y^2) + 2xy + (-16x + 6x) + (16y - 6y) + (16 - 9) = 0 \) \( 7x^2 + 7y^2 + 2xy - 10x + 10y + 7 = 0 \). This is the required equation of the conic section. To determine if it's an ellipse, we check the eccentricity. Since the eccentricity \( e = \frac{1}{2} \) is less than 1 (i.e., \( e < 1 \)), the conic section is indeed an ellipse.In simple words: We use the rule that for any point on a conic, its distance from a special point (focus) is a constant fraction ('e', the eccentricity) of its distance from a special line (directrix). We set up this equation using the given focus, directrix, and 'e'. After simplifying by squaring both sides, we get the final equation. Since the value of 'e' is \( \frac{1}{2} \), which is less than 1, we can confirm that this conic section is an ellipse.

🎯 Exam Tip: Always state whether the conic section is an ellipse, parabola, or hyperbola based on the eccentricity value ( \( e < 1 \) for ellipse, \( e = 1 \) for parabola, \( e > 1 \) for hyperbola). Be precise with algebraic steps, especially when squaring trinomials like \( (x - y - 3)^2 \).

Question 25. Find the equation of the ellipse whose foci are \( (-1, 5) \) and \( (5, 5) \) and whose major axis is 10.
Answer: The foci of the ellipse are given as \( S(-1, 5) \) and \( S'(5, 5) \). Since the y-coordinates of both foci are the same (5), the major axis of the ellipse is parallel to the x-axis. The center \( C(h, k) \) of the ellipse is the midpoint of the segment connecting the foci: \( h = \frac{-1 + 5}{2} = \frac{4}{2} = 2 \) \( k = \frac{5 + 5}{2} = \frac{10}{2} = 5 \) So, the center of the ellipse is \( (2, 5) \). The distance between the foci is \( 2ae \). \( 2ae = \sqrt{(5 - (-1))^2 + (5 - 5)^2} = \sqrt{(6)^2 + 0^2} = \sqrt{36} = 6 \). So, \( 2ae = 6 \implies ae = 3 \). The length of the major axis is given as 10. We know that the length of the major axis is \( 2a \). So, \( 2a = 10 \implies a = 5 \). Now we have \( a = 5 \) and \( ae = 3 \). We can find the eccentricity \( e \): \( 5e = 3 \implies e = \frac{3}{5} \). Since \( e = \frac{3}{5} < 1 \), this confirms it is an ellipse. Now, find \( b^2 \) using the relation \( b^2 = a^2(1 - e^2) \): \( b^2 = 5^2\left(1 - \left(\frac{3}{5}\right)^2\right) \) \( b^2 = 25\left(1 - \frac{9}{25}\right) \) \( b^2 = 25\left(\frac{25 - 9}{25}\right) = 25\left(\frac{16}{25}\right) \) \( b^2 = 16 \). The equation of an ellipse with its center at \( (h, k) \) and major axis parallel to the x-axis is: \( \frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1 \) Substitute \( h = 2, k = 5, a^2 = 25, \) and \( b^2 = 16 \): \( \frac{(x - 2)^2}{25} + \frac{(y - 5)^2}{16} = 1 \). This is the required equation of the ellipse.In simple words: We are given the locations of the two special points (foci) and the total length of the ellipse's long side (major axis). From the foci, we find the middle point, which is the center of the ellipse. We also use the distance between foci to find 'ae', and the major axis length gives us 'a'. With 'a' and 'ae', we calculate the eccentricity 'e'. Finally, we use 'a', 'e', and the center coordinates to find 'b' (the semi-minor axis) and then write the complete equation of the ellipse.

🎯 Exam Tip: When foci have the same x-coordinate, the major axis is vertical; if y-coordinates are the same, it's horizontal. The center is always the midpoint of the foci. Use \( 2a \) for major axis length and \( 2ae \) for distance between foci to find \( a \) and \( e \) efficiently.

Question 26. Find the ellipse if its foci are \( (\pm 2, 0) \) and the length of the latus rectum is \( \frac{10}{3} \).
Answer: The foci of the ellipse are given as \( (\pm 2, 0) \). This means the center of the ellipse is at the origin \( (0, 0) \), and the major axis lies along the x-axis. From the foci, we have \( ae = 2 \) ...(1) The length of the latus rectum is given as \( \frac{10}{3} \). The formula for the latus rectum is \( \frac{2b^2}{a} \). So, \( \frac{2b^2}{a} = \frac{10}{3} \). This implies \( 6b^2 = 10a \), which simplifies to \( b^2 = \frac{10a}{6} = \frac{5a}{3} \) ...(2) We also know the fundamental relation for an ellipse: \( b^2 = a^2(1 - e^2) \). Substitute \( b^2 = \frac{5a}{3} \) from (2): \( \frac{5a}{3} = a^2(1 - e^2) \) Since \( a \) cannot be 0, we can divide both sides by \( a \): \( \frac{5}{3} = a(1 - e^2) \) We know \( e = \frac{2}{a} \) from (1). Substitute this into the equation: \( \frac{5}{3} = a\left(1 - \left(\frac{2}{a}\right)^2\right) \) \( \frac{5}{3} = a\left(1 - \frac{4}{a^2}\right) \) \( \frac{5}{3} = a\left(\frac{a^2 - 4}{a^2}\right) \) \( \frac{5}{3} = \frac{a^2 - 4}{a} \) Multiply both sides by \( 3a \): \( 5a = 3(a^2 - 4) \) \( 5a = 3a^2 - 12 \) Rearrange into a quadratic equation: \( 3a^2 - 5a - 12 = 0 \) Solve this quadratic equation for \( a \). We can use factorization or the quadratic formula. Using factorization: We need two numbers that multiply to \( 3 \times -12 = -36 \) and add to -5. These numbers are -9 and 4. \( 3a^2 - 9a + 4a - 12 = 0 \) \( 3a(a - 3) + 4(a - 3) = 0 \) \( (3a + 4)(a - 3) = 0 \) This gives two possible values for \( a \): \( a = -\frac{4}{3} \) or \( a = 3 \). Since \( a \) represents a length, it must be positive. So, \( a = 3 \). Now find \( b^2 \) using equation (2): \( b^2 = \frac{5a}{3} = \frac{5 \times 3}{3} = 5 \). The equation of an ellipse with its center at \( (0, 0) \) and major axis along the x-axis is: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) Substitute \( a^2 = 3^2 = 9 \) and \( b^2 = 5 \): \( \frac{x^2}{9} + \frac{y^2}{5} = 1 \). This is the required equation of the ellipse.In simple words: We are given the locations of the two special points (foci) and the length of a specific chord (latus rectum) of an ellipse. From the foci, we know the ellipse is centered at the origin and its 'a' and 'e' values. We use the formula for the latus rectum to get 'b' in terms of 'a'. By combining these with the main ellipse formula, we form a quadratic equation for 'a'. After solving for 'a', we find 'b' and then write the final standard equation of the ellipse.

🎯 Exam Tip: When solving for 'a' from a quadratic equation, always discard any negative solutions since 'a' represents a length and must be positive. Ensure all derived values are consistent with the properties of an ellipse (e.g., \( e < 1 \)).

Question 27. Find the eccentricity of the ellipse of minor axis 2b, if the line segment joining the foci subtends an angle \( 2\alpha \) at the upper vertex. Also, find the equation of the ellipse.
Answer: Let the equation of the ellipse be \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). The minor axis length is \( 2b \). The foci are \( S'(-ae, 0) \) and \( S(ae, 0) \). The upper vertex (of the minor axis) is \( B(0, b) \). The line segment joining the foci \( S'S \) subtends an angle \( 2\alpha \) at \( B \). This means \( \angle S'BS = 2\alpha \). Consider the triangle \( \triangle SBO \), where \( O \) is the origin \( (0, 0) \). This is a right-angled triangle. The angle \( \angle SBO \) is \( \alpha \). In \( \triangle SBO \): Opposite side to \( \alpha \) is \( OS = ae \). Adjacent side to \( \alpha \) is \( OB = b \). So, \( \tan \alpha = \frac{OS}{OB} = \frac{ae}{b} \). From this, we get \( ae = b \tan \alpha \) ...(1) We also know the fundamental relation for an ellipse: \( b^2 = a^2(1 - e^2) \). From (1), \( e = \frac{b \tan \alpha}{a} \). Substitute this into the relation: \( b^2 = a^2\left(1 - \left(\frac{b \tan \alpha}{a}\right)^2\right) \) \( b^2 = a^2\left(1 - \frac{b^2 \tan^2 \alpha}{a^2}\right) \) \( b^2 = a^2 - b^2 \tan^2 \alpha \) Move \( b^2 \tan^2 \alpha \) to the left side: \( b^2 + b^2 \tan^2 \alpha = a^2 \) Factor out \( b^2 \): \( b^2(1 + \tan^2 \alpha) = a^2 \) Since \( 1 + \tan^2 \alpha = \sec^2 \alpha \): \( b^2 \sec^2 \alpha = a^2 \) Taking the square root (since \( a, b \) are positive lengths, and \( \sec \alpha \) is positive for angles in the first quadrant, typical for such geometry): \( a = b \sec \alpha \) ...(2) Now, substitute \( a = b \sec \alpha \) back into the expression for \( e \): \( e = \frac{b \tan \alpha}{a} = \frac{b \tan \alpha}{b \sec \alpha} = \frac{\sin \alpha / \cos \alpha}{1 / \cos \alpha} = \sin \alpha \). So, the eccentricity of the ellipse is \( e = \sin \alpha \). Now, to find the equation of the ellipse, we use \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Substitute \( a = b \sec \alpha \) (or \( a^2 = b^2 \sec^2 \alpha \)): \( \frac{x^2}{b^2 \sec^2 \alpha} + \frac{y^2}{b^2} = 1 \) To simplify, multiply the entire equation by \( b^2 \sec^2 \alpha \): \( x^2 + y^2 \sec^2 \alpha = b^2 \sec^2 \alpha \) This can also be written as: \( x^2 + \frac{y^2}{\cos^2 \alpha} = \frac{b^2}{\cos^2 \alpha} \) Multiply by \( \cos^2 \alpha \): \( x^2 \cos^2 \alpha + y^2 = b^2 \). This is the required equation of the ellipse.In simple words: We are given that the minor axis is \( 2b \) and the angle formed by connecting the two special points (foci) to the top end of the minor axis is \( 2\alpha \). We use trigonometry within a right triangle formed by the focus, the origin, and the minor axis endpoint to relate 'a', 'b', and 'e' to \( \tan \alpha \). Then, using the fundamental ellipse formula \( b^2 = a^2(1 - e^2) \), we solve for the eccentricity 'e', which turns out to be \( \sin \alpha \). Finally, we substitute the relationships found back into the standard equation of the ellipse to get its equation in terms of 'b' and \( \alpha \).

🎯 Exam Tip: Geometric properties like angles subtended at vertices or foci are often best approached by sketching the ellipse and using basic trigonometry (SOH CAH TOA) in right-angled triangles formed by the center, focus, and vertex. Remember \( 1 + \tan^2 \alpha = \sec^2 \alpha \) for simplification.