Question

A hyperbola having transverse axis of length 2 sin P is confocal with the ellipse 3x2 + 4y2 = 12 .Then its equation is

(A) x2 cosec2θ - y2 sec2θ = 1

(B) x2 sec2θ - y2 cosec2θ = 1

(C) x2 sin2θ - y2 cos2θ = 1

(D) x2 cos2θ - y2 sin2θ = 1

Answer 1

Correct option**(C) x2 sin2θ - y2 cos2θ = 1**

Explanation :

The equation of the ellipse is

x2/4 + y2/3 = 1

and its eccentricity is given by

3/4 = 1 - e2 ⇒ e = 1/2

Hence, the foci are (±ae,0) = (±1,0). Now, let the hyperbola be

x2/a2 - y2/b2 = 1

so that a = sin θ and the eccentricity ea is given by

b2 = a2(e'2 - 1) = sin2θ (e'2 - 1) ....(1)

Also

ae' = 1 ⇒ e' = cosecθ (:> a = sin θ)

Therefore, from Eq. (1)

b2 = sin2θ(cosec2θ - 1) = 1 - sin2 θ = cos2θ

Therefore, the equation of the hyperbola is

x2/sin2θ - y2/cos2θ = 1