Question

The area of the triangle formed by the tangents from the point $(3,2)$ to the hyperbola $x^2-9y^2=9$ and the chord of contact with respect to the point $(3,2)$ is

Answer 1

$x^2-9y^2=9$
$\Rightarrow \frac{x^2}{9}-y^2=1$
Any point on the hyperbola is $(3\sec \theta ,\tan \theta )$
Tangent at this point is
$3\sec \theta x-9\tan \theta y=9$
$\sec \theta x-3\tan \theta y=3$
This tangent passes through $(3,2)$
$3\sec \theta -6\tan \theta =3$
$\sec \theta -2\tan \theta =1$
${\sec }^2\theta =1+4\tan \theta +4{\tan }^2\theta$
$1+{\tan }^2\theta =1+4\tan \theta +4{\tan }^2\theta$
$3{\tan }^2\theta +4\tan \theta =0$
$\tan \theta =0,\tan \theta =\frac{-4}{3}$
for $\tan \theta =0$, $\sec \theta =1$
and for $\tan \theta =\frac{-4}{3}$, $\sec \theta =\frac{-5}{3}$ ($\because$ point $R$ lies in third quadrant)
$A=\frac{1}{2}\left|\begin{array}{ccc}3& 2& 1\\ 3& 0& 1\\ -5& \frac{-4}{3}& 1\end{array}\right|$
$=\frac{1}{2}\left|3\right(0+\frac{4}{3})-2(3+5)+1(-4+0\left)\right|$
$=\frac{1}{2}|4-16-4|$
$=8$