Question

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

• A. $45^{o}$
• B. $60^{o}$
• C. $30^{o}$
• D. $75^{o}$

Answer 1

Answer: (C)

$30^{o}$

Answer 2

Here $x = 2\sec\varphi$ and $y = 3\tan\varphi$

Differentiating w.r.t. φ

$\frac{dx}{d\varphi} = 2\sec\varphi\tan\varphi$ and $\frac{dy}{d\varphi} = 3\sec^{2}\varphi$

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

$\therefore$ $\frac{dy}{dx} = \frac{3}{2}\text{cosec}\varphi$ .....(i)

But tangent is parallel to $3x - y + 4 = 0$; $\therefore$ Gradient $m = 3$

......(ii)

From (i) and (ii), $\frac{3}{2}\text{cosec}\varphi = 3$ ⇒ $\text{cosec}\varphi = 2$, $\therefore$ $\varphi = 30^{o}$