Question

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

Answer 1

Point $A(\sec \theta ,2\tan \theta )$ lies on the hyperbola $2x^2-y^2=2.$
$\Rightarrow 2{\sec }^2\theta -4{\tan }^2\theta =2$
$\Rightarrow 2+2{\tan }^2\theta -4{\tan }^2\theta =2$
$\Rightarrow {\tan }^2\theta =0$
$\Rightarrow \theta =\pi n,n\in Z$

Similarly, point $B(\sec \varphi ,2\tan \varphi )$ lies on the hyperbola $2x^2-y^2=2.$
$\Rightarrow \varphi =\pi n,n\in Z$

But in question, it is given that $\theta +\varphi =\frac{\pi }{2},$ which is not possible.
Hence, it is a Bonus question.