Q: Let P (x_1, y_1 ) and Q (x_2, y_2 ) be two distinct points on the ellipse x^2 9 + y^2 4 =1 such that y_1>0, and y_2>0. Let C denote the circle x^2+y^2=9, and M be the point (3,0). Suppose the line x=x

aligned & P (3 30^ , 2 30^ ) ( 3 3 2 , 1 ) \\ & Q (3 60^ , 2 60^ ) ( 3 2 , 3 ) \\ & R ( 3 3 2 , 3 2 ), S ( 3 2 , 3 3 2 ) aligned Slope of PQ = m _ PQ = 3 -1 3 2 - 3 3 2 =- 2 3 Equation of line PQ y - 3 =- 2 3 ( x - 3 2 ) 2 x+3 y=3( 3 +1) option (A) is correct Now aligned & if N _2= ( x _2, 0 )= ( 3 2 , 0 ) \\ & | N _2 Q |= 3 and | N _2 ~S |= 3 3 2 aligned 3 | ~N _2 Q |=2 | ~N _2 ~S | option (C) is correct Now, if N _1= ( x _1, 0 ) N _1= ( 3 3 2 , 0 ) | N _1 P |=1, | ~N _1 R |= 3 2 option (D) is

Q: Consider the ellipse x 2 4 + y 2 3 = 1 . Let H α , 0 , 0 < α < 2 , be a point. A straight line drawn through H parallel to the y -axis crosses the ellip

Given x 2 4 + y 2 3 = 1 Let α ≡ 2 cos ϕ Tangent at E 2 cos ϕ , 3 sin ϕ to the ellipse is x cos ϕ 2 + y sin ϕ 3 = 1 This intersect x -axis at G 2 sec ϕ , 0 Drawing the figure as per the information, we get, Area of triangle F G H = 1 2 2 sec ϕ - 2 cos ϕ 2 sin ϕ Δ = 2 sin 2 ϕ . tan ϕ Δ = 1 - cos 2 ϕ · tan ϕ I. If ϕ = &#960

Question 1

Consider the ellipse E given by x^2 18 + y^2 12 = 1. Let H be the hyperbola whose eccentricity is the reciprocal of the eccentricity of E and whose foci are the same as that of E. Let P and Q be the p

Accepted Answer

For the ellipse E: x^2 18 + y^2 12 = 1, we have a_E^2 = 18 and b_E^2 = 12. The eccentricity e_E is given by: e_E = 1 - b_E^2 a_E^2 = 1 - 12 18 = 1 3 The foci of the ellipse E are ( a_E e_E, 0) = ( 18 1 3 , 0 ) = ( 6 , 0). For the hyperbola H, the eccentricity e_H is the reciprocal of e_E, so e_H = 3 . Since H has the same foci as E, its foci are ( 6 , 0). Let the equation of H be x^2 A^2 - y^2 B^2 = 1. The foci are ( A e_H, 0), which gives A 3 = 6 A = 2 A^2 = 2. Also, B^2 = A^2(e_H^2 - 1) = 2(3

Question 2

Passage: Consider the ellipses given by x^2 + 4y^2 = 1 and 4x^2 + y^2 = 1. Question: Let P be the point in the first quadrant where the given ellipses intersect. If is the acute angle between the tang

Accepted Answer

The given equations of the ellipses are x^2 + 4y^2 = 1 and 4x^2 + y^2 = 1. To find the point of intersection, we equate the two expressions: x^2 + 4y^2 = 4x^2 + y^2 3x^2 = 3y^2 x = y (since P lies in the first quadrant) Substituting x = y into the first ellipse equation: x^2 + 4x^2 = 1 5x^2 = 1 x = 1 5 Thus, the point of intersection is P ( 1 5 , 1 5 ). Differentiating x^2 + 4y^2 = 1 with respect to x gives the slope of the tangent to the first ellipse: 2x + 8y dy dx = 0 dy dx = - x 4y At P ( 1

Question 3

Let P (x_1, y_1 ) and Q (x_2, y_2 ) be two distinct points on the ellipse x^2 9 + y^2 4 =1 such that y_1>0, and y_2>0. Let C denote the circle x^2+y^2=9, and M be the point (3,0). Suppose the line x=x

Accepted Answer

aligned & P (3 30^ , 2 30^ ) ( 3 3 2 , 1 ) \ & Q (3 60^ , 2 60^ ) ( 3 2 , 3 ) \ & R ( 3 3 2 , 3 2 ), S ( 3 2 , 3 3 2 ) aligned Slope of PQ = m _ PQ = 3 -1 3 2 - 3 3 2 =- 2 3 Equation of line PQ y - 3 =- 2 3 ( x - 3 2 ) 2 x+3 y=3( 3 +1) option (A) is correct Now aligned & if N _2= ( x _2, 0 )= ( 3 2 , 0 ) \ & | N _2 Q |= 3 and | N _2 ~S |= 3 3 2 aligned 3 | ~N _2 Q |=2 | ~N _2 ~S | option (C) is correct Now, if N _1= ( x _1, 0 ) N _1= ( 3 3 2 , 0 ) | N _1 P |=1, | ~N _1 R |= 3 2 option (D) is

Question 4

Consider the ellipse x^2 9 + y^2 4 =1. Let S(p, q) be a point in the first quadrant such that p^2 9 + q^2 4 >1. Two tangents are drawn from S to the ellipse, of which one meets the ellipse at one end

Accepted Answer

Ar ( ORT )= 3 2 | 1 2 3 2 |= 3 2 = 1 2 = 11 6 as point T is in third quadrant T ( 3 3 2 ,-1 ) Tanget at (0,2) x(0) 9 + y(2) 4 =1 y=2...(1) Tangent at ( 3 3 2 ,-1 ) x ( 3 3 2 ) 9 + y(-1) 4 =1...(2) By solving (1) & (2) p =3 3 , q =2 Option (1) is Correct.

Question 5

Let T 1  and T 2  be two distinct common tangents to the ellipse E : x 2 6 + y 2 3 = 1  and the parabola P : y 2 = 12 x . Suppose that the tangent  T 1  touches P and E at the

Accepted Answer

Given, Equation of ellipse E : x 2 6 + y 2 3 = 1 , Now equation of tangent with slope m 1  will be : T 1 : &#160; y = m 1 x &#177; 6 m 1 2 + 3 And equation of parabola, P : y 2 = 12 x , So, equation of tangent with slope m 2  will be: y = m 2 x + 3 m 2 Now for common tangent m = m 1 = m 2 , &#160; &#177; 6 m 1 2 + 3 = 3 m 2 &#8658; &#160; m = &#177; 1 Hence, equation of common tangents will be, y = x + 3 and y = &#8211; x - 3 Now we know that, Point of contact for parabola is a m 2 , &

Question 6

Consider the ellipse  x 2 4 + y 2 3 = 1 . Let  H &#945; , 0 , 0 &#60; &#945; &#60; 2 , be a point. A straight line drawn through  H  parallel to the  y -axis crosses the ellip

Accepted Answer

Given  x 2 4 + y 2 3 = 1 Let  &#945; &#8801; 2 cos &#981; Tangent at  E 2 cos &#981; , 3 sin &#981;  to the ellipse is   x cos &#981; 2 + y sin &#981; 3 = 1 This intersect  x -axis at  G 2 sec &#981; , 0 Drawing the figure as per the information, we get,  Area of triangle  F G H = 1 2 2 sec &#981; - 2 cos &#981; 2 sin &#981; &#916; = 2 sin 2 &#981; . tan &#981; &#916; = 1 - cos 2 &#981; &#183; tan &#981; I.  If  &#981; = &#960

Question 7

Let  E  be the ellipse  x 2 16 + y 2 9 = 1 . For any three distinct points  P , Q  and  Q '  on  E , let  M P , Q  be the mid-point of the line se

Accepted Answer

. In a triangle sides joining the mid-points of two sides is half and parallel to third side M 1 M 2 = 1 2 Q Q ' Maximum value of Q Q '  is A A ' Hence, maximum value of M 1 M 2 = 1 2 A A ' = 4

Question 8

Define the collections E 1 , &#8201; E 2 , &#8201; E 3 , . . . . of ellipses and R 1 , &#8201; R 2 , &#8201; R 3 , . . . . of rectangles as follows: E 1 : x 2 9 + y 2 4 = 1 ; R 1 : rectangle of larges

Accepted Answer

As given: E 1 = x 2 a 2 + y 2 b 2 = 1 (Here a = 3 and b = 2 ) Semi major axis = a and semi minor axis = b Let a vertex of R 1 be a c o s θ , b s i n θ Then area of R 1 = 2 a c o s θ × 2 b s i n θ R 1 = 2 a b s i n 2 θ Maximum area of R 1 = 2 a b , when s i n 2 θ = 1 or, 2 θ = π 2 θ = π 4 Maximum area of R 1 = 2 a b ....(i) Now, ellipse E 2 will have semi major axis a 2 and semi minor axis b 2 ∵ E 2 = x 2 a 2 2 + y 2 b 2 2 = 1 , hence maximum area of R 2 = 2 a 2 b 2 Similarly, ellipse E 3 will ha

Ellipse — Previous Year Questions & Solutions