Question

Let $A(2 \,\sec \,\theta,\, 3 \,\tan \,\theta)$ and $B(2 \,\sec \,\phi, \,3 \,\tan \,\phi)$ where $\theta+\phi=\frac{\pi}{2}$, be two points on the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{9}=1 .$ If $(\alpha, \beta)$ is the point of intersection of normals to the hyperbola at $A$ and $B$, then $\beta$ is equal to

• A. $-\frac{13}{3}$
• B. $\frac{13}{3}$
• C. $\frac{3}{13}$
• D. $-\frac{3}{13}$

Answer 1

Answer: (A)

The normal at point $A(2 \,\sec\, \theta, \,3 \,\tan \,\theta)$ is
$ 2 \,x \,\cos \,\theta+3 \,y \cot \theta=4+9 $
$\Rightarrow 2 \,x \,\cos \,\theta+3 \,y \,\cot \,\theta=13\,\,\,...(i)$
And the normal at point $B(2 \sec \phi, 3 \tan \phi)$ is
$2 \,x \,\cos \,\phi+3 \,y \,\cot \,\phi=4+9$
$\Rightarrow 2 \,x \,\cos \,\phi+3 \,y \cot\, \phi=13\,\,\,...(ii)$
Multiplying Eq. (i) by $\cos \phi$ and Eq. (ii) by $\cos \theta$, then subtracting Eq. (ii) from Eq. (i), we get
$3\, y \,\cot \,\theta \,\cos \,\phi-3 \,y \cot \,\phi \,\cos \,\theta=13(\cos \,\phi-\cos \,\theta) $
$\Rightarrow 3 y\left\{\cot \left(\frac{\pi}{2}-\phi\right) \cos \,\phi-\cot \,\phi \,\cos \left(\frac{\pi}{2}-\phi\right)\right\} $
$=13\left\{\cos \,\phi-\cos \left(\frac{\pi}{2}-\phi\right)\right\}$
$\Rightarrow 3 \,y\{\tan\, \phi \,\cos\, \phi-\cot \,\phi\,\sin \,\phi\}=13(\cos \,\phi-\sin \,\phi)$
$\Rightarrow 3\, y(\sin \,\phi-\cos \,\phi)=13(\cos \,\phi-\sin \,\phi)$
$\Rightarrow 3\, y=\frac{-13(\sin \,\phi-\cos \,\phi)}{\sin \,\phi-\cos \,\phi}$
$\therefore y=-\frac{13}{3}$
Hence, the value of $\beta$ is $-\frac{13}{3}$.