Let A


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'.

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