Question

Let $A(A\sec \theta ,3\tan \theta )$ and $B(A\sec \varphi ,3\tan \varphi )$ 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 $\beta =$

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

Answer 1

The correct option is A $\frac{-13}{3}$

equation of hyperbola at point $A(2\sec \theta ,3\tan \theta )$ is

$y+\frac{2}{3}\sin \theta x=\frac{13}{3}\tan \theta$ ------- $\left(i\right)$

and at point $B(2sec\varphi ,3\tan \varphi )$ is

$y+\frac{2}{3}\sin \varphi x=\frac{13}{3}\tan \varphi$

now

putting $\varphi =\frac{\pi }{2}-\theta$

$y+\frac{2}{3}\cos \theta x=\frac{13}{3}\cot \theta$ ----- $\left(ii\right)$

now multiplying eq.(i) with $\cos \theta$ and eq (ii) with $\sin \theta$

then subtract both equation we find value of $\beta =-\frac{13}{3}$