Q: Let an ellipse x^2 a^2 + y^2 b^2 = 1, a < b, pass through the point (4, 3) and have eccentricity 5 3 . Then the length of its latus rectum is :

Given the equation of the ellipse is x^2 a^2 + y^2 b^2 = 1 with a The eccentricity is given by e = 5 3 . Since a Substituting the value of e: a^2 = b^2 (1 - 5 9 ) = 4 9 b^2 The ellipse passes through the point (4, 3), so substituting x = 4 and y = 3 into the equation of the ellipse gives: 16 a^2 + 9 b^2 = 1 Substituting a^2 = 4 9 b^2 into the above equation: 16 4 9 b^2 + 9 b^2 = 1 36 b^2 + 9 b^2 = 1 45 b^2 = 1 b^2 = 45 Now, finding a^2: a^2 = 4 9 45 = 20 For an ellipse with a Substituting the va

Q: An ellipse has its center at (1,-2), one focus at (3,-2) and one vertex at (5,-2). Then the length of its latus rectum is :

Center at (1, -2), focus at (3, -2), vertex at (5, -2) lie on horizontal line. Distance from center to focus: c = 2. Distance from center to vertex: a = 4. Since b^2 = a^2 - c^2 = 16 - 4 = 12. Length of latus rectum: L = 2b^2 a = 2(12) 4 = 6

Q: Let the ellipse E : x^ 2 144 + y^ 2 169 =1 and the hyperbola H : x^ 2 16 - y^ 2 ^ 2 =-1 have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H, the

Ellipse E: x^2 144 + y^2 169 = 1 has a^2 = 169, b^2 = 144, so c_E = 25 = 5 with foci at (0, 5). Hyperbola H: y^2 x^2 - x^2 16 = 1 has the same foci, so c_H^2 = x^2 + 16 = 25, giving x^2 = 9. Thus H is y^2 9 - x^2 16 = 1 with a = 3, b = 4, c = 5. Eccentricity: e = 5 3 . Latus rectum: L = 2b^2 a = 32 3 . Therefore 24(e + L) = 24 ( 5 3 + 32 3 ) = 24 37 3 = 296

Question 1

Let x^2 f(a^2+7a+3) + y^2 f(3a+15) = 1 represent an ellipse with major axis along y-axis, where f is a strictly decreasing positive function on R . If the set of all possible values of a is R - [ , ],

Accepted Answer

For the given equation to represent an ellipse with its major axis along the y-axis, the denominator of y^2 must be strictly greater than the denominator of x^2. f(3a+15) > f(a^2+7a+3) Since f is given as a strictly decreasing positive function on R , the inequality sign reverses for the arguments: 3a+15 a^2 + 4a - 12 > 0 (a+6)(a-2) > 0 The solution to this inequality is a (- , -6) (2, ). This can be rewritten in terms of the set difference as a R - [-6, 2]. Comparing this with the given set R -

Question 2

The eccentricity of an ellipse E with centre at the origin O is 3 2 and its directrices are x = 4 6 3 . Let H: x^2 a^2 - y^2 b^2 = 1 be a hyperbola whose eccentricity is equal to the length of semi-ma

Accepted Answer

For the ellipse E, the eccentricity is e_E = 3 2 and the directrices are x = a_E e_E = 4 6 3 . The semi-major axis a_E is given by: a_E = e_E 4 6 3 = 3 2 4 6 3 = 12 2 6 = 2 2 The semi-minor axis b_E is given by: b_E^2 = a_E^2(1 - e_E^2) = (2 2 )^2 (1 - 3 4 ) = 8 1 4 = 2 b_E = 2 For the hyperbola H: x^2 a^2 - y^2 b^2 = 1, its eccentricity e_H is equal to the semi-major axis of E: e_H = a_E = 2 2 The length of the latus rectum of H is equal to the length of the minor axis of E (2b_E): 2b^2 a = 2b_

Question 3

Let x = 9 be a directrix of an ellipse E, whose centre is at the origin and eccentricity is 1 3 . Let P( , 0), > 0, be a focus of E and AB be a chord passing through P. Then the locus of the mid point

Accepted Answer

Given the directrix of the ellipse is x = a e = 9 and eccentricity e = 1 3 . a 1/3 = 9 a = 3 The value of b^2 is given by b^2 = a^2(1 - e^2) = 9 (1 - 1 9 ) = 8. The equation of the ellipse is x^2 9 + y^2 8 = 1. The focus P( , 0) for > 0 is at (ae, 0) = (3 1 3 , 0 ) = (1, 0). Let the midpoint of the chord AB be (h, k). The equation of the chord in terms of its midpoint is given by T = S_1 : hx 9 + ky 8 = h^2 9 + k^2 8 Since the chord passes through the focus P(1, 0), substituting x = 1 and y = 0

Question 4

Let a focus of the ellipse E: x^2 a^2 + y^2 b^2 = 1 be S(4, 0) and its eccentricity be 4 5 . If the point P(3, ) lies on E and O is the origin, then the area of POS is equal to:

Accepted Answer

Given the focus of the ellipse S(4, 0), we have ae = 4. Since the eccentricity e = 4 5 , we get a ( 4 5 ) = 4 a = 5. Using the relation b^2 = a^2(1 - e^2), we find: b^2 = 25 (1 - 16 25 ) = 9 The equation of the ellipse is x^2 25 + y^2 9 = 1. Since the point P(3, ) lies on the ellipse, substituting x = 3 gives: 9 25 + ^2 9 = 1 ^2 9 = 1 - 9 25 = 16 25 ^2 = 144 25 | | = 12 5 The coordinates of the vertices of POS are O(0, 0), S(4, 0), and P (3, 12 5 ). The area of POS is 1 2 base height . Taking OS

Question 5

Let A be the point (3, 0) and circles with variable diameter AB touch the circle x^2 + y^2 = 36 internally. Let the curve C be the locus of the point B. If the eccentricity of C is e, then 72e^2 is eq

Accepted Answer

Let the coordinates of point B be (h, k). The center of the circle with diameter AB is C_1 ( h+3 2 , k 2 ) and its radius is r_1 = 1 2 (h-3)^2 + k^2 . The given circle is x^2 + y^2 = 36, which has center C_2(0, 0) and radius r_2 = 6. Since the circles touch internally, the distance between their centers is equal to the difference of their radii: C_1C_2 = r_2 - r_1 ( h+3 2 )^2 + ( k 2 )^2 = 6 - 1 2 (h-3)^2 + k^2 Multiplying the entire equation by 2, we get: (h+3)^2 + k^2 + (h-3)^2 + k^2 = 12 This

Question 6

Let an ellipse x^2 a^2 + y^2 b^2 = 1, a < b, pass through the point (4, 3) and have eccentricity 5 3 . Then the length of its latus rectum is :

Accepted Answer

Given the equation of the ellipse is x^2 a^2 + y^2 b^2 = 1 with a The eccentricity is given by e = 5 3 . Since a Substituting the value of e: a^2 = b^2 (1 - 5 9 ) = 4 9 b^2 The ellipse passes through the point (4, 3), so substituting x = 4 and y = 3 into the equation of the ellipse gives: 16 a^2 + 9 b^2 = 1 Substituting a^2 = 4 9 b^2 into the above equation: 16 4 9 b^2 + 9 b^2 = 1 36 b^2 + 9 b^2 = 1 45 b^2 = 1 b^2 = 45 Now, finding a^2: a^2 = 4 9 45 = 20 For an ellipse with a Substituting the va

Question 7

An ellipse has its center at (1,-2), one focus at (3,-2) and one vertex at (5,-2). Then the length of its latus rectum is :

Accepted Answer

Center at (1, -2), focus at (3, -2), vertex at (5, -2) lie on horizontal line. Distance from center to focus: c = 2. Distance from center to vertex: a = 4. Since b^2 = a^2 - c^2 = 16 - 4 = 12. Length of latus rectum: L = 2b^2 a = 2(12) 4 = 6

Question 8

Let the ellipse E : x^ 2 144 + y^ 2 169 =1 and the hyperbola H : x^ 2 16 - y^ 2 ^ 2 =-1 have the same foci. If e and L respectively denote the eccentricity and the length of the latus rectum of H, the

Accepted Answer

Ellipse E: x^2 144 + y^2 169 = 1 has a^2 = 169, b^2 = 144, so c_E = 25 = 5 with foci at (0, 5). Hyperbola H: y^2 x^2 - x^2 16 = 1 has the same foci, so c_H^2 = x^2 + 16 = 25, giving x^2 = 9. Thus H is y^2 9 - x^2 16 = 1 with a = 3, b = 4, c = 5. Eccentricity: e = 5 3 . Latus rectum: L = 2b^2 a = 32 3 . Therefore 24(e + L) = 24 ( 5 3 + 32 3 ) = 24 37 3 = 296

Question 9

Let the length of the latus rectum of an ellipse x^ 2 a^ 2 + y^ 2 b^ 2 =1,(a>b), be 30. If its eccentricity is the maximum value of the function f(t)=- 3 4 +2 t-t^ 2 , then (a^ 2 +b^ 2 ) is equal to

Accepted Answer

f(t) = -t^2 + 2t - 3 4 = -(t-1)^2 + 1 4 . Maximum value = 1 4 , so e = 1 4 . b^2 = a^2(1 - e^2) = a^2 (1 - 1 16 ) = 15a^2 16 . Latus rectum = 2b^2 a = 2 15a^2/16 a = 15a 8 = 30 a = 16. b^2 = 15 256 16 = 240. a^2 + b^2 = 256 + 240 = 496.

Question 10

Let (h, k) lie on the circle C : x^ 2 +y^ 2 =4 and the point (2 h+1,3 k+2) lie on an ellipse with eccentricity e. Then the value of 5 e^ 2 is equal to \_\_\_\_.

Accepted Answer

Since (h, k) lies on circle x^2 + y^2 = 4, we have h = 2 and k = 2 . The transformed point is (2h+1, 3k+2) = (4 + 1, 6 + 2). Rearranging: (x-1)^2 16 + (y-2)^2 36 = 1 This is an ellipse with a^2 = 36, b^2 = 16, so e^2 = 1 - b^2 a^2 = 1 - 16 36 = 5 9 Therefore 5 e^2 = 5 5/9 = 9

Question 11

Let each of the two ellipses E _ 1 : x^ 2 a^ 2 + y^ 2 b^ 2 =1,(a>b) and E _ 2 : x^ 2 ~A ^ 2 + y^ 2 ~B ^ 2 =1,( ~A < B ) have eccentricity 4 5 . Let the lengths of the latus recta of E_ 1 and E_ 2 be l

Accepted Answer

Both ellipses have e = 4/5. For E_1 (a > b): b^2 = a^2(1-e^2) = 9a^2/25. Latus rectum l_1 = 2b^2/a = 18a/25. Distance between foci of E_1: 2ae = 8a/5 = 8 a = 5, so l_1 = 18/5. For E_2 (A A^2 = B^2(1-e^2) = 9B^2/25. Latus rectum l_2 = 2A^2/B = 18B/25. Given 2l_1^2 = 9l_2: 2 324 25 = 9 18B 25 648 = 162B B = 4. Distance between foci of E_2 = 2Be = 2 4 4 5 = 32 5

Question 12

Let the line y-x=1 intersect the ellipse x^ 2 2 + y^ 2 1 =1 at the points A and B. Then the angle made by the line segment AB at the center of the ellipse is :

Accepted Answer

By solving line & equation of ellipse we get x = 0 & x = - 4 3 B (- 4 3 , - 1 3 ) m_ OB = = 1 4 AOB = 2 + = 2 + ^ -1 1 4

Question 13

Let S and S ^ be the foci of the ellipse x^ 2 25 + y^ 2 9 =1 and P ( , ) be a point on the ellipse in the first quadrant. If ( SP )^ 2 + ( S ^ P )^ 2 - SP S ^ P =37, then ^ 2 + ^ 2 is equal to :

Accepted Answer

For the ellipse x^2 25 + y^2 9 = 1, we have a = 5, b = 3, c = 4. So foci are at S( 4, 0). For point P on the ellipse: SP + S'P = 10. Given (SP)^2 + (S'P)^2 - SP S'P = 37 Let r_1 = SP and r_2 = S'P. From (r_1 + r_2)^2 = 100, we get r_1^2 + r_2^2 = 100 - 2r_1r_2. Substituting into the given equation: 100 - 3r_1r_2 = 37, so r_1r_2 = 21. Thus r_1 = 3, r_2 = 7. From ( + 4)^2 + ^2 = 9 and ( - 4)^2 + ^2 = 49, subtracting gives 16 = -40, so = - 5 2 . From the ellipse equation: ^2 = 27 4 . Therefore ^2 +

Question 14

If the line x+4 y= 7 , where R , touches the ellipse 3 x^ 2 +4 y^ 2 =1 at the point P in the first quadrant, then one of the focal distances of P is :

Accepted Answer

Ellipse 3x^2 + 4y^2 = 1: a^2 = 1 3 , b^2 = 1 4 , e = 1 2 . Tangent at (x_0, y_0): 3x_0x + 4y_0y = 1. Comparing with x + 4y = 7 : 4y_0 = 4 7 y_0 = 1 7 and 3x_0 = 7 . Substituting in ellipse equation: ^2 21 + 4 7 = 1 ^2 = 9 = 3. P = ( 1 7 , 1 7 ). Focal distances = a ex_0 = 1 3 1 2 7 . One focal distance is 1 3 + 1 2 7 .

Ellipse — Previous Year Questions & Solutions