Question

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

(a) x²cosec²θ – y²sec²θ = 1

(b) x²sec²θ – y²cosec²θ = 1

(c) x²sin²θ – y²cos²θ = 1

(d) x²cos²θ – y²sin²θ = 1

Answer 1

 The correct option (a) x²cosec² θ – y²sec² θ = 1

Explanation:

The length of transverse axis = 2sinθ = 2a

∴   a = sinθ 

Also for ellipse, 3x² + 4y² = 12

i.e.   (x²/4) + (y²/3) = 1

∴   a² = 4  and  b² = 3

Now    e = √[1 – (b²/a²)]

∴  e = √[1 – (3/4)] = ± (1/2)

∴ Focus of ellipse is = (ae, 0) & (– ae, 0)

∴ Focus = (1, 0) and (– 1, 0).

As the hyperbola if cofcal 

⇒ focus is same.

and length of transverse axis = 2sinθ 

∴ length of semi transverse axis = sinθ

i.e.   A = sinθ

and C = 1 where A, B, C are parameters in hyperbola similar to ellipse.

∴    C² = A² + B² 

∴     B² = 1 – sin²θ = cos²θ

∴ Equation of hyperbola is (x²/A²) – (y²/B²) = 1

i.e.  [x²/(sin²θ)] – [y²/(cos²θ)] = 1

i.e.     x²cosec²θ – y²sec²θ = 1.