Samacheer Kalvi Class 12 Maths Solutions Chapter 5 Two Dimensional Analytical Geometry II Exercise 5.2

Question 1. Find the equation of the parabola in each of the cases given below:
(i) focus (4, 0) and directrix x = -4.
(ii) passes through (2, – 3) and symmetric about y axis.
(iii) vertex (1,-2) and focus (4, – 2).
(iv) end points of latus rectum (4, -8) and (4, 8).

Answer:
(i) We are given the focus at \( (4, 0) \) and the directrix as \( x = -4 \). The vertex of the parabola is exactly midway between the focus and the directrix. Since the focus is \( (4,0) \) and the directrix is \( x = -4 \), the vertex is at \( (0,0) \). This parabola opens to the right, and its axis is the x-axis. The distance from the vertex to the focus is \( a \). Here, \( a = 4 \). The standard equation for a right-opening parabola with vertex \( (0,0) \) is \( y^2 = 4ax \). Substituting \( a = 4 \), we get \( y^2 = 4(4)x \). So, the equation of the parabola is \( y^2 = 16x \). This equation shows how all points on the parabola are equidistant from the focus and the directrix.
(ii) The parabola passes through the point \( (2, -3) \) and is symmetric about the y-axis. Since it is symmetric about the y-axis, its equation will be of the form \( x^2 = 4ay \). The point \( (2, -3) \) lies on the parabola, so we can substitute these coordinates into the equation: \( (2)^2 = 4a(-3) \) \( 4 = -12a \) Now, we solve for \( a \): \( a = \frac{4}{-12} = -\frac{1}{3} \) Therefore, \( 4a = -\frac{4}{3} \). Substitute the value of \( 4a \) back into the standard equation: \( x^2 = -\frac{4}{3}y \) To remove the fraction, multiply both sides by 3: \( 3x^2 = -4y \). This parabola opens downwards because the coefficient of y is negative.
(iii) We are given the vertex at \( (1, -2) \) and the focus at \( (4, -2) \). The y-coordinate of the vertex and focus is the same, which means the axis of the parabola is parallel to the x-axis. Since the focus \( (4, -2) \) is to the right of the vertex \( (1, -2) \), the parabola opens to the right. The distance \( a \) from the vertex \( (h, k) \) to the focus \( (h+a, k) \) is \( a = 4 - 1 = 3 \). The standard equation for a right-opening parabola with vertex \( (h, k) \) is \( (y - k)^2 = 4a(x - h) \). Substitute \( h = 1 \), \( k = -2 \), and \( a = 3 \) into the equation: \( (y - (-2))^2 = 4(3)(x - 1) \) \( (y + 2)^2 = 12(x - 1) \). This equation describes a parabola shifted from the origin.
(iv) We are given the end points of the latus rectum as \( (4, -8) \) and \( (4, 8) \). The latus rectum is a line segment that passes through the focus, is perpendicular to the axis of the parabola, and has its endpoints on the parabola. Since the x-coordinates of the endpoints are the same \( (x=4) \), the latus rectum is a vertical line. This means the axis of the parabola is the x-axis, and the parabola opens either to the right or left. The focus lies at the midpoint of the latus rectum. So, the focus is \( (4, \frac{8 + (-8)}{2}) = (4, 0) \). The length of the latus rectum is the distance between \( (4, -8) \) and \( (4, 8) \), which is \( |8 - (-8)| = 16 \). For a parabola, the length of the latus rectum is \( 4a \). So, \( 4a = 16 \), which means \( a = 4 \). Since the focus \( (4,0) \) is on the positive x-axis and the vertex is at the origin \( (0,0) \), the parabola opens to the right. The equation of a right-opening parabola with vertex \( (0,0) \) and focus \( (a,0) \) is \( y^2 = 4ax \). Substitute \( a = 4 \): \( y^2 = 4(4)x \) So, the equation of the parabola is \( y^2 = 16x \). The vertex is at \( (0,0) \) and the directrix is \( x = -4 \).
In simple words: For a parabola, the focus is a special point and the directrix is a special line. The vertex is in the middle of these. We use these points to find the unique equation that describes the curved shape of the parabola.

🎯 Exam Tip: Always sketch a quick diagram if coordinates are given. It helps visualize the orientation (left/right/up/down) and the vertex position of the conic section, preventing common errors.

Question 2. Find the equation of the ellipse in each of the cases given below:
(i) foci (± 3, 0), e = \( \frac{1}{2} \)
(ii) foci (0, ±4) and end points of major axis are (0, ±5).
(iii) length of latus rectum 8, eccentricity = \( \frac{3}{5} \) and major axis on x-axis.
(iv) length of latus rectum 4, distance between foci \( 4\sqrt{2} \) and major axis as y axis.

Answer:
(i) We are given the foci at \( (\pm 3, 0) \) and eccentricity \( e = \frac{1}{2} \). Comparing the foci with \( (\pm c, 0) \), we find \( c = 3 \). The eccentricity \( e \) is defined as \( e = \frac{c}{a} \). So, \( \frac{1}{2} = \frac{3}{a} \). This means \( a = 6 \). Then, \( a^2 = 6^2 = 36 \). For an ellipse, the relationship between \( a \), \( b \), and \( c \) is \( b^2 = a^2 - c^2 \). Substitute the values of \( a^2 \) and \( c^2 = 3^2 = 9 \): \( b^2 = 36 - 9 = 27 \). Since the foci are on the x-axis, the major axis is along the x-axis. The standard equation for an ellipse with its center at the origin and major axis along the x-axis is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Substitute \( a^2 = 36 \) and \( b^2 = 27 \): \( \frac{x^2}{36} + \frac{y^2}{27} = 1 \). This equation shows how the shape of the ellipse is determined by its foci and eccentricity.
(ii) We are given the foci at \( (0, \pm 4) \) and the end points of the major axis as \( (0, \pm 5) \). Comparing the foci with \( (0, \pm c) \), we find \( c = 4 \). Comparing the end points of the major axis with \( (0, \pm a) \), we find \( a = 5 \). For an ellipse, the relationship is \( b^2 = a^2 - c^2 \). Substitute \( a^2 = 5^2 = 25 \) and \( c^2 = 4^2 = 16 \): \( b^2 = 25 - 16 = 9 \). Since the foci and major axis end points are on the y-axis, the major axis is along the y-axis. The standard equation for an ellipse with its center at the origin and major axis along the y-axis is \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \). Substitute \( a^2 = 25 \) and \( b^2 = 9 \): \( \frac{x^2}{9} + \frac{y^2}{25} = 1 \). This form helps determine the shape's elongation along either axis.
(iii) We are given that the length of the latus rectum is 8 and the eccentricity \( e = \frac{3}{5} \). The major axis is on the x-axis. The formula for the length of the latus rectum for an ellipse is \( \frac{2b^2}{a} \). So, \( \frac{2b^2}{a} = 8 \). This means \( b^2 = 4a \). The relationship between \( a, b, c, \) and \( e \) for an ellipse is \( b^2 = a^2(1 - e^2) \). Substitute \( b^2 = 4a \) and \( e = \frac{3}{5} \): \( 4a = a^2(1 - (\frac{3}{5})^2) \) \( 4a = a^2(1 - \frac{9}{25}) \) \( 4a = a^2(\frac{25 - 9}{25}) \) \( 4a = a^2(\frac{16}{25}) \) Since \( a \) cannot be zero (otherwise, no ellipse exists), we can divide both sides by \( a \): \( 4 = a(\frac{16}{25}) \) To find \( a \), multiply by \( \frac{25}{16} \): \( a = 4 \times \frac{25}{16} = \frac{100}{16} = \frac{25}{4} \) Now find \( a^2 \): \( a^2 = (\frac{25}{4})^2 = \frac{625}{16} \). Next, find \( b^2 \) using \( b^2 = 4a \): \( b^2 = 4 \times \frac{25}{4} = 25 \). Since the major axis is on the x-axis, the standard equation for the ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Substitute \( a^2 = \frac{625}{16} \) and \( b^2 = 25 \): \( \frac{x^2}{\frac{625}{16}} + \frac{y^2}{25} = 1 \) This simplifies to \( \frac{16x^2}{625} + \frac{y^2}{25} = 1 \). This highlights how eccentricity impacts the length of the latus rectum.
(iv) We are given that the length of the latus rectum is 4 and the distance between foci is \( 4\sqrt{2} \). The major axis is along the y-axis. The length of the latus rectum is \( \frac{2b^2}{a} \). So, \( \frac{2b^2}{a} = 4 \). This means \( b^2 = 2a \). The distance between foci is \( 2c \). So, \( 2c = 4\sqrt{2} \). This means \( c = 2\sqrt{2} \). For an ellipse, the relationship is \( c^2 = a^2 - b^2 \). Substitute \( c = 2\sqrt{2} \) and \( b^2 = 2a \): \( (2\sqrt{2})^2 = a^2 - 2a \) \( 8 = a^2 - 2a \) Rearrange the equation into a quadratic form: \( a^2 - 2a - 8 = 0 \) Factor the quadratic equation: \( (a - 4)(a + 2) = 0 \) This gives two possible values for \( a \): \( a = 4 \) or \( a = -2 \). Since \( a \) represents a distance, it must be positive. So, \( a = 4 \). Then, \( a^2 = 4^2 = 16 \). Now find \( b^2 \) using \( b^2 = 2a \): \( b^2 = 2(4) = 8 \). Since the major axis is along the y-axis, the standard equation for the ellipse is \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \). Substitute \( a^2 = 16 \) and \( b^2 = 8 \): \( \frac{x^2}{8} + \frac{y^2}{16} = 1 \). This reveals how the distance between foci and latus rectum define the ellipse's shape.
In simple words: An ellipse is a stretched circle, defined by two points called foci. The equation changes based on whether the longer stretch (major axis) is horizontal or vertical. We use given details like foci or latus rectum length to find the unique equation for each ellipse.

🎯 Exam Tip: Remember the basic relationships for ellipses: \( c^2 = a^2 - b^2 \) (for major axis along x-axis, and \( c^2 = b^2 - a^2 \) if major axis along y-axis), and \( e = \frac{c}{a} \) (or \( e = \frac{c}{b} \) for major axis along y-axis), and length of latus rectum \( \frac{2b^2}{a} \) (or \( \frac{2a^2}{b} \) for major axis along y-axis). Be careful with the major axis direction. Here the solution used 'a' as major axis, so when major axis is along y-axis, the equation is \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \). In some textbooks, 'a' is always defined as the larger semi-axis, irrespective of its direction, which would mean it's consistent. Just ensure you consistently use the correct values of 'a' and 'b' for the denominator of the \( x^2 \) and \( y^2 \) terms according to the major axis.

Question 3. Find the equation of the hyperbola in each of the cases given below:
(i) foci (± 2, 0), eccentricity = \( \frac{3}{2} \)
(ii) centre (2, 1) one of the foci (8, 1) and corresponding directrix x = 4.
(iii) passing through (5, -2) and length of the transverse axis along x axis and length 8 units.

Answer:
(i) We are given the foci at \( (\pm 2, 0) \) and eccentricity \( e = \frac{3}{2} \). Comparing the foci with \( (\pm c, 0) \), we find \( c = 2 \). The eccentricity \( e \) is defined as \( e = \frac{c}{a} \). So, \( \frac{3}{2} = \frac{2}{a} \). This means \( 3a = 4 \), so \( a = \frac{4}{3} \). Then, \( a^2 = (\frac{4}{3})^2 = \frac{16}{9} \). For a hyperbola, the relationship between \( a \), \( b \), and \( c \) is \( b^2 = c^2 - a^2 \). Substitute the values of \( c^2 = 2^2 = 4 \) and \( a^2 = \frac{16}{9} \): \( b^2 = 4 - \frac{16}{9} = \frac{36 - 16}{9} = \frac{20}{9} \). Since the foci are on the x-axis, the transverse axis is along the x-axis. The standard equation for a hyperbola with its center at the origin and transverse axis along the x-axis is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Substitute \( a^2 = \frac{16}{9} \) and \( b^2 = \frac{20}{9} \): \( \frac{x^2}{\frac{16}{9}} - \frac{y^2}{\frac{20}{9}} = 1 \) This simplifies to \( \frac{9x^2}{16} - \frac{9y^2}{20} = 1 \). This equation demonstrates the shape of a hyperbola.
(ii) We are given the center \( (2, 1) \), one of the foci at \( (8, 1) \), and the corresponding directrix \( x = 4 \). The center of the hyperbola is \( (h, k) = (2, 1) \). One focus is \( (h+c, k) = (8, 1) \). So, \( 2+c = 8 \), which means \( c = 6 \). The distance from the center to the focus is \( ae = c \), so \( ae = 6 \). (Equation 1) The equation of the directrix corresponding to the focus \( (h+c, k) \) is \( x = h + \frac{a}{e} \). So, \( 4 = 2 + \frac{a}{e} \). This means \( \frac{a}{e} = 2 \). (Equation 2) Now we have two equations: 1) \( ae = 6 \) 2) \( \frac{a}{e} = 2 \) Multiply Equation 1 by Equation 2: \( (ae) \times (\frac{a}{e}) = 6 \times 2 \) \( a^2 = 12 \). To find \( e \), divide Equation 1 by Equation 2: \( \frac{ae}{\frac{a}{e}} = \frac{6}{2} \) \( e^2 = 3 \). Now use the relationship \( b^2 = c^2 - a^2 \). We know \( c = 6 \), so \( c^2 = 36 \). \( b^2 = 36 - 12 = 24 \). Since the focus and center have the same y-coordinate, the transverse axis is parallel to the x-axis. The standard equation for a hyperbola with center \( (h, k) \) and transverse axis parallel to the x-axis is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). Substitute \( h = 2 \), \( k = 1 \), \( a^2 = 12 \), and \( b^2 = 24 \): \( \frac{(x-2)^2}{12} - \frac{(y-1)^2}{24} = 1 \). This equation defines the hyperbola with its specified shifted center.
(iii) We are given that the hyperbola passes through \( (5, -2) \), and the length of the transverse axis along the x-axis is 8 units. The transverse axis is along the x-axis, and its length is \( 2a = 8 \). So, \( a = 4 \), which means \( a^2 = 16 \). Since the transverse axis is along the x-axis and the problem doesn't specify a shifted center, we assume the center is at the origin \( (0,0) \). The standard equation for such a hyperbola is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Substitute \( a^2 = 16 \): \( \frac{x^2}{16} - \frac{y^2}{b^2} = 1 \). The hyperbola passes through the point \( (5, -2) \), so we substitute these coordinates into the equation: \( \frac{5^2}{16} - \frac{(-2)^2}{b^2} = 1 \) \( \frac{25}{16} - \frac{4}{b^2} = 1 \) Now, we solve for \( b^2 \): \( \frac{4}{b^2} = \frac{25}{16} - 1 \) \( \frac{4}{b^2} = \frac{25 - 16}{16} \) \( \frac{4}{b^2} = \frac{9}{16} \) To find \( b^2 \), cross-multiply: \( 9b^2 = 4 \times 16 \) \( 9b^2 = 64 \) \( b^2 = \frac{64}{9} \). Substitute \( a^2 = 16 \) and \( b^2 = \frac{64}{9} \) back into the standard equation: \( \frac{x^2}{16} - \frac{y^2}{\frac{64}{9}} = 1 \) This simplifies to \( \frac{x^2}{16} - \frac{9y^2}{64} = 1 \). This equation helps understand how the hyperbola expands from its center.
In simple words: A hyperbola looks like two parabolas facing away from each other. Its equation depends on its center, foci, and how wide it opens. We use the given information to find the values that correctly describe its specific shape.

🎯 Exam Tip: Distinguish carefully between ellipses \( (b^2 = a^2 - c^2) \) and hyperbolas \( (b^2 = c^2 - a^2) \) when using the \( a,b,c \) relationship. The sign difference is critical. Always double-check the major/transverse axis orientation to correctly set up the standard equation.

Question 4. Find the vertex, focus, equation of directrix and length of the latus rectum of the following:
(i) \( y^2 = 16x \)
(ii) \( x^2 = 24y \)
(iii) \( y^2 = -8x \)
(iv) \( x^2 – 2x + 8y + 17 = 0 \)
(v) \( y^2 – 4y – 8x + 12 = 0 \)

Answer:
(i) Given the equation of the parabola: \( y^2 = 16x \). This is in the standard form \( y^2 = 4ax \). By comparing, we get \( 4a = 16 \), so \( a = 4 \). 1. **Vertex:** For \( y^2 = 4ax \), the vertex is at \( (0, 0) \). 2. **Focus:** For \( y^2 = 4ax \), the focus is at \( (a, 0) \). So, the focus is \( (4, 0) \). 3. **Equation of Directrix:** For \( y^2 = 4ax \), the directrix is \( x = -a \). So, the directrix is \( x = -4 \), which can also be written as \( x + 4 = 0 \). 4. **Length of Latus Rectum:** For any parabola, the length of the latus rectum is \( 4a \). So, the length is \( 4(4) = 16 \). This parabola opens to the right.
(ii) Given the equation of the parabola: \( x^2 = 24y \). This is in the standard form \( x^2 = 4ay \). By comparing, we get \( 4a = 24 \), so \( a = 6 \). 1. **Vertex:** For \( x^2 = 4ay \), the vertex is at \( (0, 0) \). 2. **Focus:** For \( x^2 = 4ay \), the focus is at \( (0, a) \). So, the focus is \( (0, 6) \). 3. **Equation of Directrix:** For \( x^2 = 4ay \), the directrix is \( y = -a \). So, the directrix is \( y = -6 \), which can also be written as \( y + 6 = 0 \). 4. **Length of Latus Rectum:** The length is \( 4a \). So, the length is \( 4(6) = 24 \). This parabola opens upwards.
(iii) Given the equation of the parabola: \( y^2 = -8x \). This is in the standard form \( y^2 = -4ax \). By comparing, we get \( -4a = -8 \), so \( a = 2 \). 1. **Vertex:** For \( y^2 = -4ax \), the vertex is at \( (0, 0) \). 2. **Focus:** For \( y^2 = -4ax \), the focus is at \( (-a, 0) \). So, the focus is \( (-2, 0) \). 3. **Equation of Directrix:** For \( y^2 = -4ax \), the directrix is \( x = a \). So, the directrix is \( x = 2 \), which can also be written as \( x - 2 = 0 \). 4. **Length of Latus Rectum:** The length is \( 4a \). So, the length is \( 4(2) = 8 \). This parabola opens to the left.
(iv) Given the equation of the parabola: \( x^2 – 2x + 8y + 17 = 0 \). To find the vertex, focus, directrix, and latus rectum, we first convert the equation to standard form by completing the square. Group the x-terms and move y-terms and constants to the right side: \( x^2 - 2x = -8y - 17 \) Complete the square for the x-terms: \( (x - 1)^2 - 1 = -8y - 17 \) \( (x - 1)^2 = -8y - 17 + 1 \) \( (x - 1)^2 = -8y - 16 \) Factor out -8 from the right side: \( (x - 1)^2 = -8(y + 2) \) This equation is in the standard form \( (x - h)^2 = -4a(y - k) \). By comparing, we find: \( h = 1 \), \( k = -2 \) \( -4a = -8 \), so \( a = 2 \). 1. **Vertex:** The vertex is \( (h, k) \). So, the vertex is \( (1, -2) \). 2. **Focus:** For this form, the focus is \( (h, k - a) \). So, the focus is \( (1, -2 - 2) = (1, -4) \). 3. **Equation of Directrix:** For this form, the directrix is \( y = k + a \). So, the directrix is \( y = -2 + 2 = 0 \). 4. **Length of Latus Rectum:** The length is \( 4a \). So, the length is \( 4(2) = 8 \). This parabola opens downwards.
(v) Given the equation of the parabola: \( y^2 – 4y – 8x + 12 = 0 \). To find the vertex, focus, directrix, and latus rectum, we first convert the equation to standard form by completing the square. Group the y-terms and move x-terms and constants to the right side: \( y^2 - 4y = 8x - 12 \) Complete the square for the y-terms: \( (y - 2)^2 - 4 = 8x - 12 \) \( (y - 2)^2 = 8x - 12 + 4 \) \( (y - 2)^2 = 8x - 8 \) Factor out 8 from the right side: \( (y - 2)^2 = 8(x - 1) \) This equation is in the standard form \( (y - k)^2 = 4a(x - h) \). By comparing, we find: \( h = 1 \), \( k = 2 \) \( 4a = 8 \), so \( a = 2 \). 1. **Vertex:** The vertex is \( (h, k) \). So, the vertex is \( (1, 2) \). 2. **Focus:** For this form, the focus is \( (h + a, k) \). So, the focus is \( (1 + 2, 2) = (3, 2) \). 3. **Equation of Directrix:** For this form, the directrix is \( x = h - a \). So, the directrix is \( x = 1 - 2 = -1 \), which can also be written as \( x + 1 = 0 \). 4. **Length of Latus Rectum:** The length is \( 4a \). So, the length is \( 4(2) = 8 \). This parabola opens to the right.
In simple words: For each parabola, we first rewrite its equation into a simpler, standard form. This helps us quickly find its key features: the vertex (the turning point), the focus (a special point), the directrix (a special line), and the latus rectum (a measure of its width).

🎯 Exam Tip: When dealing with general forms like \( x^2 – 2x + 8y + 17 = 0 \), always complete the square for the squared term (x or y) to transform it into the standard form \( (x-h)^2 = 4a(y-k) \) or \( (y-k)^2 = 4a(x-h) \). This makes it easy to identify the vertex \( (h,k) \) and the value of \( a \).

Question 5. Identify the type of conic and find centre, foci, vertices and directrices of each of the following:
(i) \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \)
(ii) \( \frac{x^2}{3} + \frac{y^2}{10} = 1 \)
(iii) \( \frac{x^2}{25} – \frac{y^2}{144} = 1 \)
(iv) \( \frac{y^2}{16} – \frac{x^2}{9} = 1 \)

Answer:
(i) Given the equation: \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \). **Type:** This is an ellipse because both \( x^2 \) and \( y^2 \) terms are positive and are added together. Compare with \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, \( a^2 = 25 \Rightarrow a = 5 \). And \( b^2 = 9 \Rightarrow b = 3 \). Since \( a^2 > b^2 \), the major axis is along the x-axis. Now, we find \( c \): \( c^2 = a^2 - b^2 = 25 - 9 = 16 \Rightarrow c = 4 \). 1. **Centre:** The center is at \( (0, 0) \). 2. **Foci:** The foci are at \( (\pm c, 0) \). So, the foci are \( (\pm 4, 0) \). 3. **Vertices:** The vertices are at \( (\pm a, 0) \). So, the vertices are \( (\pm 5, 0) \). 4. **Directrices:** The directrices are \( x = \pm \frac{a}{e} \). First, find eccentricity \( e = \frac{c}{a} = \frac{4}{5} \). Then \( \frac{a}{e} = \frac{5}{\frac{4}{5}} = \frac{25}{4} \). So, the directrices are \( x = \pm \frac{25}{4} \). This shows how the fundamental values define the ellipse's specific layout.
(ii) Given the equation: \( \frac{x^2}{3} + \frac{y^2}{10} = 1 \). **Type:** This is an ellipse because both \( x^2 \) and \( y^2 \) terms are positive and are added together. Compare with \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) (since \( 10 > 3 \), \( a^2 = 10 \)). Here, \( a^2 = 10 \Rightarrow a = \sqrt{10} \). And \( b^2 = 3 \Rightarrow b = \sqrt{3} \). Since \( a^2 > b^2 \), the major axis is along the y-axis. Now, we find \( c \): \( c^2 = a^2 - b^2 = 10 - 3 = 7 \Rightarrow c = \sqrt{7} \). 1. **Centre:** The center is at \( (0, 0) \). 2. **Foci:** The foci are at \( (0, \pm c) \). So, the foci are \( (0, \pm \sqrt{7}) \). 3. **Vertices:** The vertices are at \( (0, \pm a) \). So, the vertices are \( (0, \pm \sqrt{10}) \). 4. **Directrices:** The directrices are \( y = \pm \frac{a}{e} \). First, find eccentricity \( e = \frac{c}{a} = \frac{\sqrt{7}}{\sqrt{10}} \). Then \( \frac{a}{e} = \frac{\sqrt{10}}{\frac{\sqrt{7}}{\sqrt{10}}} = \frac{10}{\sqrt{7}} \). So, the directrices are \( y = \pm \frac{10}{\sqrt{7}} \). This shows the vertical alignment of the ellipse.
In simple words: Identifying conics means figuring out if an equation describes a circle, ellipse, parabola, or hyperbola. Each has special points like the center, foci, and vertices, and defining lines called directrices. We use the numbers in the equation to find all these specific features.

🎯 Exam Tip: To identify the type of conic, look at the signs between the \( x^2 \) and \( y^2 \) terms in the general equation: if they are both positive and coefficients are different, it's an ellipse. If one is positive and one is negative, it's a hyperbola. For ellipses, 'a' is always the larger semi-axis, so adjust the standard equation (\( x^2/a^2 \) or \( y^2/a^2 \)) accordingly based on which denominator is larger. If 'a' and 'b' are used as defined by the variable, then just apply the rules consistently.

Question 5. conic and find centre, foci, vertices and directrices of each of the following:
(iii) \( \frac {x^2}{25} – \frac {y^2}{144} = 1 \)
Answer: The given equation \( \frac {x^2}{25} - \frac {y^2}{144} = 1 \) represents a hyperbola. For this hyperbola, the transverse axis is along the x-axis, meaning it opens horizontally.
From the equation, we have \( a^2 = 25 \) and \( b^2 = 144 \).
So, \( a = 5 \) and \( b = 12 \).
We calculate \( c^2 = a^2 + b^2 = 25 + 144 = 169 \), which means \( c = 13 \).
The eccentricity \( e = \frac {c}{a} = \frac {13}{5} \).

(a) Centre: The center of this hyperbola is at \( (0, 0) \).
(b) Vertices: The vertices are at \( (\pm a, 0) = (\pm 5, 0) \).
(c) Foci: The foci are at \( (\pm c, 0) = (\pm 13, 0) \).
(d) Equation of the directrices: The directrices are given by \( x = \pm \frac {a}{e} = \pm \frac {5}{\frac{13}{5}} = \pm \frac {25}{13} \).
This conic's equation provides all its key properties directly from the standard form.
In simple words: This shape is a hyperbola that opens sideways. Its center is at (0,0). The points where it curves outwards are at (±5,0), and its focal points are further out at (±13,0). The guide lines (directrices) for this hyperbola are at \( x = \pm \frac {25}{13} \).

🎯 Exam Tip: Remember that for a hyperbola, \( c^2 = a^2 + b^2 \), unlike an ellipse where \( c^2 = a^2 - b^2 \). This is a common mistake.

Question 5. conic and find centre, foci, vertices and directrices of each of the following:
(iv) \( \frac {y^2}{16} – \frac {x^2}{9} = 1 \)
Answer: The equation \( \frac {y^2}{16} - \frac {x^2}{9} = 1 \) represents a hyperbola. In this case, since \( y^2 \) is the positive term, the transverse axis is along the y-axis, meaning the hyperbola opens vertically.
From the equation, we have \( a^2 = 16 \) and \( b^2 = 9 \).
Therefore, \( a = 4 \) and \( b = 3 \).
We calculate \( c^2 = a^2 + b^2 = 16 + 9 = 25 \), so \( c = 5 \).
The eccentricity \( e = \frac {c}{a} = \frac {5}{4} \).

(a) Centre: The center of this hyperbola is at \( (0, 0) \).
(b) Vertices: The vertices are at \( (0, \pm a) = (0, \pm 4) \).
(c) Foci: The foci are at \( (0, \pm c) = (0, \pm 5) \).
(d) Equation of the directrices: The directrices are given by \( y = \pm \frac {a}{e} = \pm \frac {4}{\frac{5}{4}} = \pm \frac {16}{5} \).
This type of hyperbola, opening along the y-axis, is often referred to as a vertical hyperbola.
In simple words: This is a hyperbola that opens up and down. Its center is at (0,0). The points where it turns are at (0,±4), and its special focal points are at (0,±5). The guide lines for this hyperbola are at \( y = \pm \frac {16}{5} \).

🎯 Exam Tip: For hyperbolas, if the \( y^2 \) term is positive, the transverse axis is vertical. If the \( x^2 \) term is positive, it's horizontal. This determines the orientation of the branches.

Question 6. Prove that the length of the latus rectum of the hyperbola \( \frac {x^2}{a^2} - \frac {y^2}{b^2} = 1 \) is \( \frac {2b^2}{a} \)
Answer: To prove the length of the latus rectum for a hyperbola, we consider a point on the latus rectum. The latus rectum is a line segment passing through a focus and perpendicular to the transverse axis.
For the hyperbola \( \frac {x^2}{a^2} - \frac {y^2}{b^2} = 1 \), a focus is at \( S(ae, 0) \).
Let \( L(ae, y_1) \) be one endpoint of the latus rectum in the first quadrant. Since L lies on the hyperbola, its coordinates must satisfy the equation:
\( \frac {(ae)^2}{a^2} - \frac {y_1^2}{b^2} = 1 \)
\( \frac {a^2e^2}{a^2} - \frac {y_1^2}{b^2} = 1 \)
\( e^2 - \frac {y_1^2}{b^2} = 1 \)
We know that for a hyperbola, \( e^2 = 1 + \frac {b^2}{a^2} \).
Substitute this into the equation:
\( (1 + \frac {b^2}{a^2}) - \frac {y_1^2}{b^2} = 1 \)
\( 1 + \frac {b^2}{a^2} - \frac {y_1^2}{b^2} = 1 \)
\( \frac {b^2}{a^2} = \frac {y_1^2}{b^2} \)
\( y_1^2 = \frac {b^4}{a^2} \)
Taking the square root, we get \( y_1 = \pm \frac {b^2}{a} \).
So, the coordinates of the endpoints of the latus rectum are \( (ae, \frac {b^2}{a}) \) and \( (ae, -\frac {b^2}{a}) \).
The length of the latus rectum \( LL' \) is the distance between these two points, which is \( \frac {b^2}{a} - (-\frac {b^2}{a}) = \frac {b^2}{a} + \frac {b^2}{a} = \frac {2b^2}{a} \).
Hence, the length of the latus rectum is proven to be \( \frac {2b^2}{a} \). This formula helps determine the 'width' of the hyperbola at its foci.
In simple words: The latus rectum is a special line across the hyperbola at its focus. If you measure its total length, it will always be equal to \( \frac {2b^2}{a} \). This is found by plugging a focus's x-coordinate into the hyperbola's equation.

🎯 Exam Tip: Remember the definition of the latus rectum and how to use the hyperbola's equation and eccentricity formula to derive its length.

Question 7. Show that the absolute value of the difference of the focal distances of any point P on the hyperbola is the length of its transverse axis.
Answer: This property is a fundamental definition of a hyperbola. Let P be any point on the hyperbola, and let \( S \) and \( S' \) be the two foci of the hyperbola.
By the definition of a conic section, for any point P on the hyperbola, the ratio of its distance from a focus to its distance from the corresponding directrix is equal to the eccentricity \( e \).
So, for focus \( S \) and its directrix, \( SP = e \cdot PM \), where \( PM \) is the perpendicular distance from P to the directrix.
Similarly, for focus \( S' \) and its directrix, \( S'P = e \cdot PM' \), where \( PM' \) is the perpendicular distance from P to the other directrix.
Let the equations of the directrices be \( x = \frac {a}{e} \) and \( x = -\frac {a}{e} \). For a point \( P(x_0, y_0) \) on the hyperbola, assume it is on the right branch.
Then \( PM = x_0 - \frac {a}{e} \) and \( PM' = x_0 + \frac {a}{e} \).
So, \( SP = e \left( x_0 - \frac {a}{e} \right) = ex_0 - a \).
And \( S'P = e \left( x_0 + \frac {a}{e} \right) = ex_0 + a \).
The absolute value of the difference of the focal distances is \( |S'P - SP| \).
\( |S'P - SP| = |(ex_0 + a) - (ex_0 - a)| = |ex_0 + a - ex_0 + a| = |2a| \).
Since \( a \) is a positive length, \( |2a| = 2a \).
The length of the transverse axis of a hyperbola is \( 2a \). Therefore, the absolute value of the difference of the focal distances of any point P on the hyperbola is equal to the length of its transverse axis. This constant difference is what defines a hyperbola's shape.
In simple words: Imagine you pick any point on a hyperbola. Now, measure its distance to one special point (focus S) and its distance to another special point (focus S'). If you subtract these two distances, you will always get the same number, and that number is exactly the length of the hyperbola's main axis, called the transverse axis.

🎯 Exam Tip: Clearly state the definition of focal distances using eccentricity and directrices. Ensure your algebraic steps lead to \( |2a| \), the length of the transverse axis.

Question 8. Identify the type of conic and find centre, foci, vertices and directrices of each of the following:
(i) \( \frac {(x-3)^2}{225} + \frac {(y-4)^2}{289} = 1 \)
Answer: The given equation is in the form \( \frac {(x-h)^2}{b^2} + \frac {(y-k)^2}{a^2} = 1 \). This indicates that the conic is an ellipse, and its major axis is parallel to the y-axis because \( a^2 \) is under \( (y-k)^2 \).
From the equation, \( a^2 = 289 \) and \( b^2 = 225 \).
So, \( a = 17 \) and \( b = 15 \).
We calculate \( c^2 = a^2 - b^2 = 289 - 225 = 64 \), which gives \( c = 8 \).
The eccentricity is \( e = \frac {c}{a} = \frac {8}{17} \).

(a) Centre: The center of the ellipse \( (h, k) \) is \( (3, 4) \).
(b) Vertices: The vertices are \( (h, k \pm a) = (3, 4 \pm 17) \). This gives two vertices: \( (3, 4+17) = (3, 21) \) and \( (3, 4-17) = (3, -13) \).
(c) Foci: The foci are \( (h, k \pm c) = (3, 4 \pm 8) \). This gives two foci: \( (3, 4+8) = (3, 12) \) and \( (3, 4-8) = (3, -4) \).
(d) Directrices: The equations of the directrices are \( y = k \pm \frac {a}{e} = 4 \pm \frac {17}{\frac{8}{17}} = 4 \pm \frac {289}{8} \).
This results in two directrices: \( y = \frac {32 + 289}{8} = \frac {321}{8} \) and \( y = \frac {32 - 289}{8} = \frac {-257}{8} \).
The properties of an ellipse define its precise location and shape on the coordinate plane.
In simple words: This shape is an ellipse. It is taller than it is wide, with its center at (3,4). The top and bottom points (vertices) are at (3,21) and (3,-13). Its special focal points are at (3,12) and (3,-4). The guide lines (directrices) are horizontal lines at \( y = \frac {321}{8} \) and \( y = \frac {-257}{8} \).

🎯 Exam Tip: Pay attention to which term has the larger denominator; this tells you if the major axis is horizontal or vertical. Remember to shift the coordinates by (h,k) for non-centered conics.

Question 8. Identify the type of conic and find centre, foci, vertices and directrices of each of the following:
(ii) \( \frac {(x+1)^2}{100} + \frac {(y-2)^2}{64} = 1 \)
Answer: The given equation is in the standard form for an ellipse: \( \frac {(x-h)^2}{a^2} + \frac {(y-k)^2}{b^2} = 1 \). This means it is an ellipse with its major axis parallel to the x-axis, because \( a^2 \) is under \( (x-h)^2 \).
From the equation, \( a^2 = 100 \) and \( b^2 = 64 \).
So, \( a = 10 \) and \( b = 8 \).
We calculate \( c^2 = a^2 - b^2 = 100 - 64 = 36 \), which gives \( c = 6 \).
The eccentricity is \( e = \frac {c}{a} = \frac {6}{10} = \frac {3}{5} \).

(a) Centre: The center of the ellipse \( (h, k) \) is \( (-1, 2) \).
(b) Vertices: The vertices are \( (h \pm a, k) = (-1 \pm 10, 2) \). This gives two vertices: \( (-1+10, 2) = (9, 2) \) and \( (-1-10, 2) = (-11, 2) \).
(c) Foci: The foci are \( (h \pm c, k) = (-1 \pm 6, 2) \). This gives two foci: \( (-1+6, 2) = (5, 2) \) and \( (-1-6, 2) = (-7, 2) \).
(d) Directrices: The equations of the directrices are \( x = h \pm \frac {a}{e} = -1 \pm \frac {10}{\frac{3}{5}} = -1 \pm \frac {50}{3} \).
This results in two directrices: \( x = \frac {-3 + 50}{3} = \frac {47}{3} \) and \( x = \frac {-3 - 50}{3} = \frac {-53}{3} \).
Understanding the role of 'h' and 'k' is essential for accurately locating the features of a shifted conic.
In simple words: This shape is an ellipse, wider than it is tall, with its center at (-1,2). Its far-end points (vertices) are at (9,2) and (-11,2). The special focal points are at (5,2) and (-7,2). The guide lines (directrices) are vertical lines at \( x = \frac {47}{3} \) and \( x = \frac {-53}{3} \).

🎯 Exam Tip: Always remember that the values of \( h \) and \( k \) in the standard form \( (x-h)^2 \) and \( (y-k)^2 \) determine the center of the conic, affecting all other coordinates.

Question 8. Identify the type of conic and find centre, foci, vertices and directrices of each of the following:
(iii) \( \frac {(x+3)^2}{225} - \frac {(y-4)^2}{64} = 1 \)
Answer: The given equation is in the form \( \frac {(x-h)^2}{a^2} - \frac {(y-k)^2}{b^2} = 1 \). This indicates that the conic is a hyperbola, and its transverse axis is parallel to the x-axis, meaning it opens horizontally.
From the equation, \( a^2 = 225 \) and \( b^2 = 64 \).
So, \( a = 15 \) and \( b = 8 \).
We calculate \( c^2 = a^2 + b^2 = 225 + 64 = 289 \), which gives \( c = 17 \).
The eccentricity is \( e = \frac {c}{a} = \frac {17}{15} \).

(a) Centre: The center of the hyperbola \( (h, k) \) is \( (-3, 4) \).
(b) Vertices: The vertices are \( (h \pm a, k) = (-3 \pm 15, 4) \). This gives two vertices: \( (-3+15, 4) = (12, 4) \) and \( (-3-15, 4) = (-18, 4) \).
(c) Foci: The foci are \( (h \pm c, k) = (-3 \pm 17, 4) \). This gives two foci: \( (-3+17, 4) = (14, 4) \) and \( (-3-17, 4) = (-20, 4) \).
(d) Directrices: The equations of the directrices are \( x = h \pm \frac {a}{e} = -3 \pm \frac {15}{\frac{17}{15}} = -3 \pm \frac {225}{17} \).
This results in two directrices: \( x = \frac {-51 + 225}{17} = \frac {174}{17} \) and \( x = \frac {-51 - 225}{17} = \frac {-276}{17} \).
The direction of opening and the precise location of the center are key to correctly identifying all these features.
In simple words: This shape is a hyperbola that opens sideways, with its center at (-3,4). Its turning points (vertices) are at (12,4) and (-18,4). The special focal points are at (14,4) and (-20,4). The guide lines (directrices) are vertical lines at \( x = \frac {174}{17} \) and \( x = \frac {-276}{17} \).

🎯 Exam Tip: For hyperbolas, the \( a^2 \) term always corresponds to the semi-transverse axis, regardless of whether it's smaller or larger than \( b^2 \).

Question 8. Identify the type of conic and find centre, foci, vertices and directrices of each of the following:
(iv) \( \frac {(y-2)^2}{25} - \frac {(x+1)^2}{16} = 1 \)
Answer: The given equation is in the form \( \frac {(y-k)^2}{a^2} - \frac {(x-h)^2}{b^2} = 1 \). This indicates that the conic is a hyperbola, and its transverse axis is parallel to the y-axis, meaning it opens vertically.
From the equation, \( a^2 = 25 \) and \( b^2 = 16 \).
So, \( a = 5 \) and \( b = 4 \).
We calculate \( c^2 = a^2 + b^2 = 25 + 16 = 41 \), which gives \( c = \sqrt {41} \).
The eccentricity is \( e = \frac {c}{a} = \frac {\sqrt {41}}{5} \).

(a) Centre: The center of the hyperbola \( (h, k) \) is \( (-1, 2) \).
(b) Vertices: The vertices are \( (h, k \pm a) = (-1, 2 \pm 5) \). This gives two vertices: \( (-1, 2+5) = (-1, 7) \) and \( (-1, 2-5) = (-1, -3) \).
(c) Foci: The foci are \( (h, k \pm c) = (-1, 2 \pm \sqrt {41}) \). This gives two foci: \( (-1, 2 + \sqrt {41}) \) and \( (-1, 2 - \sqrt {41}) \).
(d) Directrices: The equations of the directrices are \( y = k \pm \frac {a}{e} = 2 \pm \frac {5}{\frac{\sqrt {41}}{5}} = 2 \pm \frac {25}{\sqrt {41}} \).
This results in two directrices: \( y = 2 + \frac {25}{\sqrt {41}} \) and \( y = 2 - \frac {25}{\sqrt {41}} \).
Recognizing the leading positive term \( y^2 \) is crucial for determining the vertical orientation of the hyperbola.
In simple words: This shape is a hyperbola that opens up and down, with its center at (-1,2). Its turning points (vertices) are at (-1,7) and (-1,-3). The special focal points are at \( (-1, 2 \pm \sqrt {41}) \). The guide lines (directrices) are horizontal lines at \( y = 2 \pm \frac {25}{\sqrt {41}} \).

🎯 Exam Tip: When \( c \) involves a square root, remember to express the foci and directrices using that root, as it is an exact value.

Question 8. Identify the type of conic and find centre, foci, vertices and directrices of each of the following:
(v) \( 18x^2 + 12y^2 – 144x + 48y + 120 = 0 \)
Answer: First, we need to rewrite the equation in standard form by completing the square for both x and y terms:
\( 18x^2 - 144x + 12y^2 + 48y = -120 \)
Factor out the coefficients of \( x^2 \) and \( y^2 \):
\( 18(x^2 - 8x) + 12(y^2 + 4y) = -120 \)
Complete the square inside the parentheses:
\( 18(x^2 - 8x + 16) + 12(y^2 + 4y + 4) = -120 + 18(16) + 12(4) \)
\( 18(x - 4)^2 + 12(y + 2)^2 = -120 + 288 + 48 \)
\( 18(x - 4)^2 + 12(y + 2)^2 = 216 \)
Divide by 216 to get 1 on the right side:
\( \frac {18(x-4)^2}{216} + \frac {12(y+2)^2}{216} = 1 \)
\( \frac {(x-4)^2}{12} + \frac {(y+2)^2}{18} = 1 \)
This is the standard form of an ellipse. Since \( 18 > 12 \), the major axis is parallel to the y-axis.
From the equation, \( a^2 = 18 \) and \( b^2 = 12 \).
So, \( a = \sqrt {18} = 3\sqrt {2} \) and \( b = \sqrt {12} = 2\sqrt {3} \).
We calculate \( c^2 = a^2 - b^2 = 18 - 12 = 6 \), which gives \( c = \sqrt {6} \).
The eccentricity is \( e = \frac {c}{a} = \frac {\sqrt {6}}{3\sqrt {2}} = \frac {1}{\sqrt {3}} \).

(a) Centre: The center of the ellipse \( (h, k) \) is \( (4, -2) \).
(b) Vertices: The vertices are \( (h, k \pm a) = (4, -2 \pm 3\sqrt {2}) \). This gives two vertices: \( (4, -2 + 3\sqrt {2}) \) and \( (4, -2 - 3\sqrt {2}) \).
(c) Foci: The foci are \( (h, k \pm c) = (4, -2 \pm \sqrt {6}) \). This gives two foci: \( (4, -2 + \sqrt {6}) \) and \( (4, -2 - \sqrt {6}) \).
(d) Directrices: The equations of the directrices are \( y = k \pm \frac {a}{e} = -2 \pm \frac {3\sqrt {2}}{\frac{1}{\sqrt {3}}} = -2 \pm 3\sqrt {6} \).
This results in two directrices: \( y = -2 + 3\sqrt {6} \) and \( y = -2 - 3\sqrt {6} \).
The first step in solving such problems is always to transform the general equation into its standard form.
In simple words: First, we change the complex equation into a simpler form, which shows it is an ellipse that stands tall. Its center is at (4,-2). The top and bottom points (vertices) are at \( (4, -2 \pm 3\sqrt {2}) \). The special focal points are at \( (4, -2 \pm \sqrt {6}) \). The guide lines (directrices) are horizontal lines at \( y = -2 \pm 3\sqrt {6} \).

🎯 Exam Tip: When completing the square, remember to add the adjusted constant to both sides of the equation. Also, distribute the factored-out coefficient correctly before adding.

Question 8. Identify the type of conic and find centre, foci, vertices and directrices of each of the following:
(vi) \( 9x^2 – y^2 – 36x – 6y + 18 = 0 \)
Answer: First, we need to rewrite the equation in standard form by completing the square for both x and y terms:
\( 9x^2 - 36x - y^2 - 6y = -18 \)
Factor out the coefficients of \( x^2 \) and \( y^2 \):
\( 9(x^2 - 4x) - (y^2 + 6y) = -18 \)
Complete the square inside the parentheses:
\( 9(x^2 - 4x + 4) - (y^2 + 6y + 9) = -18 + 9(4) - 1(9) \)
\( 9(x - 2)^2 - (y + 3)^2 = -18 + 36 - 9 \)
\( 9(x - 2)^2 - (y + 3)^2 = 9 \)
Divide by 9 to get 1 on the right side:
\( \frac {9(x-2)^2}{9} - \frac {(y+3)^2}{9} = 1 \)
\( \frac {(x-2)^2}{1} - \frac {(y+3)^2}{9} = 1 \)
This is the standard form of a hyperbola. Since the \( x^2 \) term is positive, the transverse axis is parallel to the x-axis, meaning it opens horizontally.
From the equation, \( a^2 = 1 \) and \( b^2 = 9 \).
So, \( a = 1 \) and \( b = 3 \).
We calculate \( c^2 = a^2 + b^2 = 1 + 9 = 10 \), which gives \( c = \sqrt {10} \).
The eccentricity is \( e = \frac {c}{a} = \frac {\sqrt {10}}{1} = \sqrt {10} \).

(a) Centre: The center of the hyperbola \( (h, k) \) is \( (2, -3) \).
(b) Vertices: The vertices are \( (h \pm a, k) = (2 \pm 1, -3) \). This gives two vertices: \( (2+1, -3) = (3, -3) \) and \( (2-1, -3) = (1, -3) \).
(c) Foci: The foci are \( (h \pm c, k) = (2 \pm \sqrt {10}, -3) \). This gives two foci: \( (2 + \sqrt {10}, -3) \) and \( (2 - \sqrt {10}, -3) \).
(d) Directrices: The equations of the directrices are \( x = h \pm \frac {a}{e} = 2 \pm \frac {1}{\sqrt {10}} \).
This results in two directrices: \( x = 2 + \frac {1}{\sqrt {10}} \) and \( x = 2 - \frac {1}{\sqrt {10}} \).
It is important to remember that for a hyperbola, the order of \( a^2 \) and \( b^2 \) in the denominator does not determine the larger value, but rather which term is positive.
In simple words: We change the complex equation into a simpler form, which reveals it's a hyperbola opening sideways. Its center is at (2,-3). The turning points (vertices) are at (3,-3) and (1,-3). The special focal points are at \( (2 \pm \sqrt {10}, -3) \). The guide lines (directrices) are vertical lines at \( x = 2 \pm \frac {1}{\sqrt {10}} \).

🎯 Exam Tip: Be cautious with the signs when completing the square, especially when a negative sign is factored out, as this can easily lead to calculation errors.