Question

From any point on the hyperbola $\frac{x^2}{4}-\frac{y^2}{1}=1$, tangents are drawn to the hyperbola $\frac{x^2}{8}-\frac{y^2}{2}=1$. The area cut-off by the chord of contact on the asymptotes is equal to

Answer 1

Let $(2\sec \theta ,\tan \theta )$ be the point on hyperbola $\frac{x^2}{4}-\frac{y^2}{1}=1$
$\therefore$ Chord of contact of tangents drawn from the point $(2\sec \theta ,\tan \theta )$ to the hyperbola $\frac{x^2}{8}-\frac{y^2}{2}=1$ is $\frac{x\sec \theta }{4}-\frac{y\tan \theta }{2}=1$
Asymptotes of the hyperbola $\frac{x^2}{8}-\frac{y^2}{2}=1$ is $\frac{x}{2\sqrt{2}}\pm \frac{y}{\sqrt{2}}=0$
Intersection points of the asymptotes to the chord of contact are
$(\frac{4\cos \theta }{1-\sin \theta },\frac{2\cos \theta }{1-\sin \theta })\text{and}(\frac{4\cos \theta }{1+\sin \theta },-\frac{2\cos \theta }{1+\sin \theta })$

So required area is $=\frac{1}{2}|x_2{y}_1-y_2{x}_1|=\frac{1}{2}|\frac{4\cos \theta }{1+\sin \theta }\times \frac{2\cos \theta }{1-\sin \theta }+\frac{2\cos \theta }{1+\sin \theta }\times \frac{4\cos \theta }{1-\sin \theta }|=8$