Question

If the tangent on the point (2 sec φ, 3 tan φ) of the hyperbola $\frac{x^{2}}{4} - \frac{y^{2}}{9}$ = 1 is parallel to 3x – y + 4 = 0, then the value of φ is

• A. 45⁰
• B. 60⁰
• C. 30⁰
• D. 75⁰

Answer 1

Answer: (C)

30⁰

Answer 2

Differentiation of x = 2 sec φ ⇒ $\frac{dx}{d\varphi}$ = 2 sec ⇒ tan ⇒

Differentiate, y = 3 tan ⇒ w.r.t ⇒, we get $\frac{dy}{d\varphi}$ = 3 sec² φ

∴ Gradient of tangent $\frac{dy}{dx} = \frac{dy/d\varphi}{dx/d\varphi} = \frac{3\sec^{2}\varphi}{2\sec\varphi\tan\varphi}$

$\frac{dy}{dx} = \frac{3}{2}$cosec φ

But, tangent is parallel to 3x – y + 4 = 0

∴ Gradient m = 3

By (I) and (ii), $\frac{3}{2}$cosec φ = 3 ⇒ cosec φ = 2, ∴ φ = 30⁰.