Let $A(\sec \theta, 2 \tan \theta)$ and $B(\sec \phi, 2 \tan \phi)$, where $\theta+\phi=\pi / 2$, be two points on the hyperbola $2 x^{2}-y^{2}=2$. 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
Since, point $A(\sec \theta, 2 \tan \theta)$ lies on the hyperbola
$2 x^{2}-y^{2}=2$
Therefore, $2 \sec ^{2} \theta-4 \tan ^{2} \theta=2$
$\Rightarrow 2+2 \tan ^{2} \theta-4 \tan ^{2} \theta=2$
$\Rightarrow \tan \theta=0 \Rightarrow \theta=0$
Similarly, for point $\mathrm{B}$, we will get $\phi=0$.
but according to question $\theta+\phi=\frac{\pi}{2}$
which is not possible.
Hence it must be a 'BONUS'.
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.