Question

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