GSEB Class 11 Maths Solutions Chapter 11 Conic Sections Exercise 11.3

In each of the following questions 1 to 9, find the co-ordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and length of latus rectum of the ellipse:

Solutions to Questions 1 to 9:

 

Question 1. Equation of ellipse is \( \frac{x^{2}}{36} + \frac{y^{2}}{16} = 1 \).
Answer: Here, we have \( a^{2} = 36 \) and \( b^{2} = 16 \). This gives us \( a = 6 \) and \( b = 4 \). We calculate \( c^{2} = a^{2} - b^{2} \), so \( c^{2} = 36 - 16 = 20 \). Therefore, \( c = \pm \sqrt{20} = \pm 2\sqrt{5} \). The eccentricity \( e \) is given by \( e = \frac{c}{a} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3} \). The co-ordinates of the foci are \( (\pm c, 0) \), which means \( (\pm 2\sqrt{5}, 0) \). The vertices are \( (\pm a, 0) \), so \( (\pm 6, 0) \). The length of the major axis is \( 2a = 2 \times 6 = 12 \). The length of the minor axis is \( 2b = 2 \times 4 = 8 \). Finally, the length of the latus rectum is \( \frac{2b^{2}}{a} = \frac{2 \times 16}{6} = \frac{16}{3} \).
In simple words: For this ellipse, we found the larger radius (a) is 6 and the smaller radius (b) is 4. The distance to the foci (c) is \( 2\sqrt{5} \). The foci are at \( (\pm 2\sqrt{5}, 0) \), and the vertices are at \( (\pm 6, 0) \). The major axis measures 12, the minor axis measures 8, and the eccentricity is \( \frac{\sqrt{5}}{3} \).

Exam Tip: Remember to identify \( a^{2} \) and \( b^{2} \) correctly from the equation, especially when the denominator under \( x^{2} \) is smaller than that under \( y^{2} \), which would mean the major axis is along the y-axis.

 

Question 2. Equation of ellipse is \( \frac{x^{2}}{4} + \frac{y^{2}}{25} = 1 \).
Answer: In this equation, \( b^{2} = 4 \), so \( b = 2 \). Also, \( a^{2} = 25 \), which means \( a = 5 \). Since \( a^{2} \) is under \( y^{2} \), the major axis lies along the y-axis. Calculate \( c^{2} = a^{2} - b^{2} = 25 - 4 = 21 \). Thus, \( c = \sqrt{21} \). The co-ordinates of the foci are \( (0, \pm c) \), i.e., \( (0, \pm \sqrt{21}) \). The vertices are \( (0, \pm a) \), which means \( (0, \pm 5) \). The length of the major axis is \( 2a = 2 \times 5 = 10 \). The length of the minor axis is \( 2b = 2 \times 2 = 4 \). The eccentricity \( e = \frac{c}{a} = \frac{\sqrt{21}}{5} \). The latus rectum's length is \( \frac{2b^{2}}{a} = \frac{2 \times 4}{5} = \frac{8}{5} \).
In simple words: This ellipse has its main axis along the y-axis because \( a^{2} \) is under \( y^{2} \). The foci are at \( (0, \pm \sqrt{21}) \), and the highest/lowest points (vertices) are at \( (0, \pm 5) \). The long axis is 10 units long, and the short axis is 4 units long.

Exam Tip: When \( a^{2} \) is under \( y^{2} \), remember that the major axis is vertical, so foci and vertices will have x-coordinate 0.

 

Question 3. Equation of ellipse is \( \frac{x^{2}}{16} + \frac{y^{2}}{9} = 1 \).
Answer: Here, \( a^{2} = 16 \), so \( a = 4 \). And \( b^{2} = 9 \), which gives \( b = 3 \). Since \( a^{2} \) is under \( x^{2} \), the major axis is along the x-axis. To find \( c \), we use \( c^{2} = a^{2} - b^{2} = 16 - 9 = 7 \). Thus, \( c = \sqrt{7} \). The co-ordinates of the foci are \( (\pm c, 0) \), which are \( (\pm \sqrt{7}, 0) \). The vertices are \( (\pm a, 0) \), i.e., \( (\pm 4, 0) \). The length of the major axis is \( 2a = 2 \times 4 = 8 \). The length of the minor axis is \( 2b = 2 \times 3 = 6 \). The eccentricity \( e = \frac{c}{a} = \frac{\sqrt{7}}{4} \). The length of the latus rectum is \( \frac{2b^{2}}{a} = \frac{2 \times 9}{4} = \frac{9}{2} \).
In simple words: For this ellipse, the main axis is along the x-axis. The foci are at \( (\pm \sqrt{7}, 0) \), and the vertices are at \( (\pm 4, 0) \). The major axis has a length of 8, and the minor axis has a length of 6.

Exam Tip: Always double-check whether \( a^{2} \) is under \( x^{2} \) or \( y^{2} \) to determine if the major axis is horizontal or vertical, respectively.

 

Question 4. Equation of ellipse is \( \frac{x^{2}}{25} + \frac{y^{2}}{100} = 1 \).
Answer: In this equation, \( a^{2} = 100 \), so \( a = 10 \). Also, \( b^{2} = 25 \), which means \( b = 5 \). Because \( a^{2} \) is under \( y^{2} \), the major axis is along the y-axis. To find \( c \), we use \( c^{2} = a^{2} - b^{2} = 100 - 25 = 75 \). Thus, \( c = \sqrt{75} = 5\sqrt{3} \). The foci are at \( (0, \pm c) \), which means \( (0, \pm 5\sqrt{3}) \). The vertices are at \( (0, \pm a) \), i.e., \( (0, \pm 10) \). The length of the major axis is \( 2a = 2 \times 10 = 20 \). The length of the minor axis is \( 2b = 2 \times 5 = 10 \). The eccentricity \( e = \frac{c}{a} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2} \). The length of the latus rectum is \( \frac{2b^{2}}{a} = \frac{2 \times 25}{10} = 5 \).
In simple words: This ellipse is taller than it is wide. Its main axis is on the y-axis. The foci are at \( (0, \pm 5\sqrt{3}) \), and the highest/lowest points (vertices) are at \( (0, \pm 10) \). The long axis is 20 units, and the short axis is 10 units.

Exam Tip: Be careful when determining the value of 'a' and 'b' from the denominators; 'a' is always associated with the larger denominator, which defines the major axis.

 

Question 5. \( \frac{x^{2}}{49} + \frac{y^{2}}{36} = 1 \) is the equation of ellipse.
Answer: Here, \( a^{2} = 49 \), so \( a = 7 \). Also, \( b^{2} = 36 \), which means \( b = 6 \). Since \( a^{2} \) is under \( x^{2} \), the major axis is along the x-axis. To find \( c \), we use \( c^{2} = a^{2} - b^{2} = 49 - 36 = 13 \). Thus, \( c = \sqrt{13} \). The co-ordinates of the foci are \( (\pm c, 0) \), which means \( (\pm \sqrt{13}, 0) \). The vertices are \( (\pm a, 0) \), i.e., \( (\pm 7, 0) \). The length of the major axis is \( 2a = 2 \times 7 = 14 \). The length of the minor axis is \( 2b = 2 \times 6 = 12 \). The eccentricity \( e = \frac{c}{a} = \frac{\sqrt{13}}{7} \). The length of the latus rectum is \( \frac{2b^{2}}{a} = \frac{2 \times 36}{7} = \frac{72}{7} \).
In simple words: For this ellipse, the main axis is on the x-axis. The foci are at \( (\pm \sqrt{13}, 0) \), and the vertices are at \( (\pm 7, 0) \). The major axis has a length of 14, and the minor axis has a length of 12.

Exam Tip: Organize your calculations for \( a, b, c, e \) and lengths clearly to avoid mistakes, especially with square roots and fractions.

 

Question 6. \( \frac{x^{2}}{100} + \frac{y^{2}}{400} = 1 \) is the equation of the ellipse.
Answer: In this equation, \( a^{2} = 400 \), so \( a = 20 \). Also, \( b^{2} = 100 \), which means \( b = 10 \). Since \( a^{2} \) is under \( y^{2} \), the major axis is along the y-axis. To find \( c \), we use \( c^{2} = a^{2} - b^{2} = 400 - 100 = 300 \). Thus, \( c = \sqrt{300} = 10\sqrt{3} \). The vertices are at \( (0, \pm a) \), i.e., \( (0, \pm 20) \). The foci are at \( (0, \pm c) \), which means \( (0, \pm 10\sqrt{3}) \). The length of the major axis is \( 2a = 2 \times 20 = 40 \). The length of the minor axis is \( 2b = 2 \times 10 = 20 \). The eccentricity \( e = \frac{c}{a} = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2} \). The length of the latus rectum is \( \frac{2b^{2}}{a} = \frac{2 \times 100}{20} = 10 \).
In simple words: This ellipse is elongated vertically. Its major axis is along the y-axis. The vertices are at \( (0, \pm 20) \), and the foci are at \( (0, \pm 10\sqrt{3}) \). The long axis is 40 units, and the short axis is 20 units.

Exam Tip: Always simplify square roots (like \( \sqrt{300} \) to \( 10\sqrt{3} \)) for the final answer to present it in the most precise form.

 

Question 7. \( 36x^{2} + 4y^{2} = 144 \) is the equation of the ellipse.
Answer: First, we need to convert the given equation into standard form by dividing all terms by 144: \( \frac{36x^{2}}{144} + \frac{4y^{2}}{144} = \frac{144}{144} \), which simplifies to \( \frac{x^{2}}{4} + \frac{y^{2}}{36} = 1 \). Here, \( a^{2} = 36 \), so \( a = 6 \). Also, \( b^{2} = 4 \), which means \( b = 2 \). Since \( a^{2} \) is under \( y^{2} \), the major axis is along the y-axis. To find \( c \), we use \( c^{2} = a^{2} - b^{2} = 36 - 4 = 32 \). Thus, \( c = \sqrt{32} = 4\sqrt{2} \). The foci are at \( (0, \pm c) \), which means \( (0, \pm 4\sqrt{2}) \). The vertices are at \( (0, \pm a) \), i.e., \( (0, \pm 6) \). The length of the major axis is \( 2a = 2 \times 6 = 12 \). The length of the minor axis is \( 2b = 2 \times 2 = 4 \). The eccentricity \( e = \frac{c}{a} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3} \). The length of the latus rectum is \( \frac{2b^{2}}{a} = \frac{2 \times 4}{6} = \frac{4}{3} \).
In simple words: First, rewrite the equation to fit the standard ellipse form. This ellipse is vertical. The foci are at \( (0, \pm 4\sqrt{2}) \), and the vertices are at \( (0, \pm 6) \). The major axis is 12 units, and the minor axis is 4 units.

Exam Tip: Always convert the given equation to the standard form \( \frac{x^{2}}{A^{2}} + \frac{y^{2}}{B^{2}} = 1 \) before finding parameters, by dividing by the constant on the right side.

 

Question 8. The equation of the ellipse is \( 16x^{2} + y^{2} = 16 \).
Answer: First, we change the given equation into standard form by dividing all terms by 16: \( \frac{16x^{2}}{16} + \frac{y^{2}}{16} = \frac{16}{16} \), which simplifies to \( \frac{x^{2}}{1} + \frac{y^{2}}{16} = 1 \). Here, \( a^{2} = 16 \), so \( a = 4 \). Also, \( b^{2} = 1 \), which means \( b = 1 \). Since \( a^{2} \) is under \( y^{2} \), the major axis is along the y-axis. To find \( c \), we use \( c^{2} = a^{2} - b^{2} = 16 - 1 = 15 \). Thus, \( c = \sqrt{15} \). The vertices are at \( (0, \pm a) \), i.e., \( (0, \pm 4) \). The length of the major axis is \( 2a = 2 \times 4 = 8 \). The length of the minor axis is \( 2b = 2 \times 1 = 2 \). The eccentricity \( e = \frac{c}{a} = \frac{\sqrt{15}}{4} \). The length of the latus rectum is \( \frac{2b^{2}}{a} = \frac{2 \times 1}{4} = \frac{1}{2} \).
In simple words: After rearranging the equation, we find this ellipse is vertical. The vertices are at \( (0, \pm 4) \). The major axis is 8 units long, and the minor axis is 2 units long.

Exam Tip: Be attentive to the coefficients of \( x^{2} \) and \( y^{2} \) in the given equation and ensure proper division to obtain the standard form.

 

Question 9. Equation of ellipse is \( 4x^{2} + 9y^{2} = 36 \).
Answer: First, we need to convert the given equation into standard form by dividing all terms by 36: \( \frac{4x^{2}}{36} + \frac{9y^{2}}{36} = \frac{36}{36} \), which simplifies to \( \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 \). Here, \( a^{2} = 9 \), so \( a = 3 \). Also, \( b^{2} = 4 \), which means \( b = 2 \). Since \( a^{2} \) is under \( x^{2} \), the major axis is along the x-axis. To find \( c \), we use \( c^{2} = a^{2} - b^{2} = 9 - 4 = 5 \). Thus, \( c = \sqrt{5} \). The co-ordinates of the foci are \( (\pm c, 0) \), which means \( (\pm \sqrt{5}, 0) \). The vertices are \( (\pm a, 0) \), i.e., \( (\pm 3, 0) \). The length of the major axis is \( 2a = 2 \times 3 = 6 \). The length of the minor axis is \( 2b = 2 \times 2 = 4 \). The eccentricity \( e = \frac{c}{a} = \frac{\sqrt{5}}{3} \). The length of the latus rectum is \( \frac{2b^{2}}{a} = \frac{2 \times 4}{3} = \frac{8}{3} \).
In simple words: After putting the equation in standard form, we see this ellipse is horizontal. The foci are at \( (\pm \sqrt{5}, 0) \), and the vertices are at \( (\pm 3, 0) \). The long axis measures 6 units, and the short axis measures 4 units.

Exam Tip: Remember that \( a^{2} \) is always the larger denominator and \( b^{2} \) is the smaller one, which helps identify the orientation of the major axis.

In each of the following questions 10 to 20, find the equation for the ellipse that satisfies the given conditions:

Solutions Questions 10-20:

 

Question 10. Vertices \( (\pm 5, 0) \); Foci \( (\pm 4, 0) \).
Answer: From the given vertices \( (\pm 5, 0) \), we know that \( (\pm a, 0) = (\pm 5, 0) \), so \( a = 5 \). From the foci \( (\pm 4, 0) \), we know that \( (\pm ae, 0) = (\pm 4, 0) \), so \( ae = 4 \). Using these values, we can find the eccentricity: \( e = \frac{4}{a} = \frac{4}{5} \). To find \( b^{2} \), we use the relation \( b^{2} = a^{2}(1 - e^{2}) \). Substituting the values, \( b^{2} = 5^{2}(1 - (\frac{4}{5})^{2}) = 25(1 - \frac{16}{25}) = 25(\frac{25 - 16}{25}) = 25(\frac{9}{25}) = 9 \). Thus, \( b = 3 \). Since the vertices and foci are on the x-axis, the major axis is along the x-axis. The equation of the ellipse is \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \). Plugging in the values, we get \( \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 \). This can also be written as \( 9x^{2} + 25y^{2} = 225 \).
In simple words: The vertices tell us \( a=5 \), and the foci tell us \( ae=4 \), which helps us find \( e=\frac{4}{5} \). Then, we calculate \( b^2=9 \). Since the foci and vertices are on the x-axis, the equation is \( \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 \).

Exam Tip: Always note the coordinates of the vertices and foci to determine the orientation of the major axis (x-axis or y-axis) first.

 

Question 11. Foci \( (0, \pm 5) \), vertices \( (0, \pm 13) \).
Answer: From the given foci \( (0, \pm 5) \), we know that \( (0, \pm ae) = (0, \pm 5) \), so \( ae = 5 \). From the vertices \( (0, \pm 13) \), we know that \( (0, \pm a) = (0, \pm 13) \), so \( a = 13 \). We can find the eccentricity: \( e = \frac{ae}{a} = \frac{5}{13} \). To find \( b^{2} \), we use the relation \( b^{2} = a^{2} - (ae)^{2} = 13^{2} - 5^{2} = 169 - 25 = 144 \). Thus, \( b = 12 \). Since the vertices and foci are on the y-axis, the major axis is along the y-axis. The equation of the ellipse is \( \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 \). Plugging in the values, we get \( \frac{x^{2}}{144} + \frac{y^{2}}{169} = 1 \).
In simple words: The given points reveal \( a=13 \) and \( ae=5 \). We then calculate \( b^2=144 \). Since the foci and vertices are on the y-axis, the equation for the ellipse is \( \frac{x^{2}}{144} + \frac{y^{2}}{169} = 1 \).

Exam Tip: If the x-coordinates of foci and vertices are zero, it means the major axis is along the y-axis, and the standard form is \( \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 \).

 

Question 12. Vertices \( (\pm 6, 0) \); Foci \( (\pm 4, 0) \).
Answer: From the given vertices \( (\pm 6, 0) \), we get \( a = 6 \). From the foci \( (\pm 4, 0) \), we get \( c = 4 \). Since the vertices and foci are on the x-axis, the major axis is along the x-axis. To find \( b^{2} \), we use the relation \( b^{2} = a^{2} - c^{2} \). Substituting the values, \( b^{2} = 6^{2} - 4^{2} = 36 - 16 = 20 \). The equation of the ellipse is \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \). Plugging in the values, we get \( \frac{x^{2}}{36} + \frac{y^{2}}{20} = 1 \).
In simple words: From the vertices, \( a=6 \). From the foci, \( c=4 \). Using these, we find \( b^2=20 \). As the main axis is on the x-axis, the ellipse's equation is \( \frac{x^{2}}{36} + \frac{y^{2}}{20} = 1 \).

Exam Tip: When both 'a' and 'c' are directly given or easily derived, using \( b^{2} = a^{2} - c^{2} \) is the most efficient way to find \( b^{2} \).

 

Question 13. Ends of major axis \( (\pm 3, 0) \) and ends of minor axis \( (0, \pm 2) \).
Answer: The ends of the major axis are \( (\pm 3, 0) \), which means \( a = 3 \). Since these points are on the x-axis, the major axis is along the x-axis. The ends of the minor axis are \( (0, \pm 2) \), which means \( b = 2 \). The equation of the ellipse is \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \). Plugging in the values \( a = 3 \) and \( b = 2 \), we get \( \frac{x^{2}}{3^{2}} + \frac{y^{2}}{2^{2}} = 1 \), which simplifies to \( \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 \).
In simple words: The ends of the major axis give us \( a=3 \) (and confirm the x-axis as the major axis). The ends of the minor axis give us \( b=2 \). The ellipse equation is then \( \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 \).

Exam Tip: The endpoints of the major axis are \( (\pm a, 0) \) or \( (0, \pm a) \), and the endpoints of the minor axis are \( (0, \pm b) \) or \( (\pm b, 0) \), always perpendicular to the major axis.

 

Question 14. Ends of major axis \( (0, \pm \sqrt{5}) \) and ends of minor axis \( (\pm 1, 0) \).
Answer: The ends of the major axis are \( (0, \pm \sqrt{5}) \), which means \( a = \sqrt{5} \). Since these points are on the y-axis, the major axis is along the y-axis. The ends of the minor axis are \( (\pm 1, 0) \), which means \( b = 1 \). The equation of the ellipse is \( \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 \). Plugging in the values \( a = \sqrt{5} \) and \( b = 1 \), we get \( \frac{x^{2}}{1^{2}} + \frac{y^{2}}{(\sqrt{5})^{2}} = 1 \), which simplifies to \( \frac{x^{2}}{1} + \frac{y^{2}}{5} = 1 \).
In simple words: The major axis points give us \( a=\sqrt{5} \) and confirm the y-axis as the major axis. The minor axis points give us \( b=1 \). So, the ellipse's equation is \( \frac{x^{2}}{1} + \frac{y^{2}}{5} = 1 \).

Exam Tip: Be careful with square roots when squaring 'a' and 'b' for the denominators; \( (\sqrt{5})^2 = 5 \).

 

Question 15. Length of major axis 26, Foci \( (\pm 5, 0) \).
Answer: The length of the major axis is 26, so \( 2a = 26 \), which means \( a = 13 \). The foci are \( (\pm 5, 0) \), so \( c = 5 \). Since the foci are on the x-axis, the major axis is along the x-axis. To find \( b^{2} \), we use the relation \( b^{2} = a^{2} - c^{2} \). Substituting the values, \( b^{2} = 13^{2} - 5^{2} = 169 - 25 = 144 \). The equation of the ellipse is \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \). Plugging in the values, we get \( \frac{x^{2}}{13^{2}} + \frac{y^{2}}{144} = 1 \), which simplifies to \( \frac{x^{2}}{169} + \frac{y^{2}}{144} = 1 \).
In simple words: We are given \( a=13 \) from the major axis length, and \( c=5 \) from the foci. We then calculate \( b^2=144 \). Since the foci are on the x-axis, the equation for the ellipse is \( \frac{x^{2}}{169} + \frac{y^{2}}{144} = 1 \).

Exam Tip: Pay attention to the distinction between 'length of major axis' (which is \( 2a \)) and 'length of semi-major axis' (which is \( a \)).

 

Question 16. Length of minor axis = 16, Foci \( (0, \pm 6) \).
Answer: The length of the minor axis is 16, so \( 2b = 16 \), which means \( b = 8 \). The foci are \( (0, \pm 6) \), so \( c = 6 \). Since the foci are on the y-axis, the major axis is along the y-axis. To find \( a^{2} \), we use the relation \( a^{2} = b^{2} + c^{2} \). Substituting the values, \( a^{2} = 8^{2} + 6^{2} = 64 + 36 = 100 \). The equation of the ellipse is \( \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 \). Plugging in the values, we get \( \frac{x^{2}}{8^{2}} + \frac{y^{2}}{100} = 1 \), which simplifies to \( \frac{x^{2}}{64} + \frac{y^{2}}{100} = 1 \).
In simple words: The minor axis length gives us \( b=8 \), and the foci give us \( c=6 \). We then calculate \( a^2=100 \). As the foci are on the y-axis, the equation for the ellipse is \( \frac{x^{2}}{64} + \frac{y^{2}}{100} = 1 \).

Exam Tip: Remember that for an ellipse, \( a^{2} \) is always the sum of \( b^{2} \) and \( c^{2} \), regardless of the major axis orientation.

 

Question 17. Foci \( (\pm 3, 0) \); \( a = 4 \).
Answer: The foci are \( (\pm 3, 0) \), which means \( c = 3 \). We are given \( a = 4 \). Since the foci are on the x-axis, the major axis is along the x-axis. To find \( b^{2} \), we use the relation \( b^{2} = a^{2} - c^{2} \). Substituting the values, \( b^{2} = 4^{2} - 3^{2} = 16 - 9 = 7 \). The equation of the ellipse is \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \). Plugging in the values, we get \( \frac{x^{2}}{4^{2}} + \frac{y^{2}}{7} = 1 \), which simplifies to \( \frac{x^{2}}{16} + \frac{y^{2}}{7} = 1 \).
In simple words: From the foci, we have \( c=3 \). We are also given \( a=4 \). We calculate \( b^2=7 \). Since the foci are on the x-axis, the ellipse's equation is \( \frac{x^{2}}{16} + \frac{y^{2}}{7} = 1 \).

Exam Tip: Always write down the known values of \( a, b, \) or \( c \) from the problem statement first to guide your next steps.

 

Question 18. \( b = 3 \), \( c = 4 \), centre at the origin, foci on x-axis.
Answer: We are given \( b = 3 \) and \( c = 4 \). Since the foci are on the x-axis, the major axis of the ellipse lies along the x-axis. We know the relationship \( a^{2} = b^{2} + c^{2} \) for an ellipse. Substituting the given values, \( a^{2} = 3^{2} + 4^{2} = 9 + 16 = 25 \). Therefore, \( a = \sqrt{25} = 5 \). The standard equation for an ellipse with its major axis along the x-axis is \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \). Plugging in the values for \( a^{2} \) and \( b^{2} \), we get \( \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 \). This is the desired equation for the ellipse.
In simple words: We are given \( b=3 \) and \( c=4 \). Because the foci are on the x-axis, the major axis is horizontal. We find \( a^2=25 \). The equation for the ellipse becomes \( \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 \).

Exam Tip: The location of the foci (x-axis or y-axis) directly determines the orientation of the ellipse and thus which standard equation to use.

 

Question 19. Centre at \( (0, 0) \), major axis on the y-axis and passes through the points \( (3, 2) \) and \( (1, 6) \).
Answer: Since the major axis is on the y-axis and the center is at \( (0, 0) \), the equation of the ellipse is \( \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 \). The ellipse passes through the points \( (3, 2) \) and \( (1, 6) \). Substitute \( (3, 2) \) into the equation: \( \frac{3^{2}}{b^{2}} + \frac{2^{2}}{a^{2}} = 1 \implies \frac{9}{b^{2}} + \frac{4}{a^{2}} = 1 \quad \dots(1) \) Substitute \( (1, 6) \) into the equation: \( \frac{1^{2}}{b^{2}} + \frac{6^{2}}{a^{2}} = 1 \implies \frac{1}{b^{2}} + \frac{36}{a^{2}} = 1 \quad \dots(2) \) To solve these simultaneous equations, multiply equation (1) by 9: \( \frac{81}{b^{2}} + \frac{36}{a^{2}} = 9 \quad \dots(3) \) Subtract equation (2) from equation (3): \( (\frac{81}{b^{2}} + \frac{36}{a^{2}}) - (\frac{1}{b^{2}} + \frac{36}{a^{2}}) = 9 - 1 \) \( \frac{80}{b^{2}} = 8 \implies 80 = 8b^{2} \implies b^{2} = 10 \). Substitute \( b^{2} = 10 \) into equation (1): \( \frac{9}{10} + \frac{4}{a^{2}} = 1 \) \( \frac{4}{a^{2}} = 1 - \frac{9}{10} = \frac{1}{10} \) \( a^{2} = 40 \). Therefore, the equation of the ellipse is \( \frac{x^{2}}{10} + \frac{y^{2}}{40} = 1 \).
In simple words: Since the major axis is vertical, we use the equation \( \frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 \). We plug in the two given points to get two equations. Solving these equations helps us find \( b^2=10 \) and \( a^2=40 \). So, the final equation is \( \frac{x^{2}}{10} + \frac{y^{2}}{40} = 1 \).

Exam Tip: When given points the ellipse passes through, substitute them into the general equation to form simultaneous equations and solve for \( a^{2} \) and \( b^{2} \).

 

Question 20. Major axis on x-axis and passes through \( (4, 3) \) and \( (6, 2) \).
Answer: Since the major axis is on the x-axis, the equation of the ellipse is \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \). The ellipse passes through the points \( (4, 3) \) and \( (6, 2) \). Substitute \( (4, 3) \) into the equation: \( \frac{4^{2}}{a^{2}} + \frac{3^{2}}{b^{2}} = 1 \implies \frac{16}{a^{2}} + \frac{9}{b^{2}} = 1 \quad \dots(1) \) Substitute \( (6, 2) \) into the equation: \( \frac{6^{2}}{a^{2}} + \frac{2^{2}}{b^{2}} = 1 \implies \frac{36}{a^{2}} + \frac{4}{b^{2}} = 1 \quad \dots(2) \) Subtract equation (2) from equation (1): \( (\frac{16}{a^{2}} + \frac{9}{b^{2}}) - (\frac{36}{a^{2}} + \frac{4}{b^{2}}) = 1 - 1 \) \( \frac{16 - 36}{a^{2}} + \frac{9 - 4}{b^{2}} = 0 \) \( \frac{-20}{a^{2}} + \frac{5}{b^{2}} = 0 \) Rearrange the equation: \( \frac{5}{b^{2}} = \frac{20}{a^{2}} \implies 5a^{2} = 20b^{2} \implies a^{2} = 4b^{2} \). Substitute \( a^{2} = 4b^{2} \) into equation (1): \( \frac{16}{4b^{2}} + \frac{9}{b^{2}} = 1 \) \( \frac{4}{b^{2}} + \frac{9}{b^{2}} = 1 \) \( \frac{13}{b^{2}} = 1 \implies b^{2} = 13 \). Now, find \( a^{2} \): \( a^{2} = 4b^{2} = 4 \times 13 = 52 \). Therefore, the equation of the ellipse is \( \frac{x^{2}}{52} + \frac{y^{2}}{13} = 1 \).
In simple words: Since the major axis is horizontal, we use \( \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \). We plug in the two given points to create two equations. Solving these allows us to find \( a^2=52 \) and \( b^2=13 \). The final equation for the ellipse is \( \frac{x^{2}}{52} + \frac{y^{2}}{13} = 1 \).

Exam Tip: Eliminating one variable (like \( a^{2} \) or \( b^{2} \)) by multiplying equations and subtracting them is a common strategy for these types of problems.