Question

A tangent drawn to hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at $P\left(\frac{\pi }{6}\right)$ forms a triangle of area $3a^{2}$ square units, with coordinate axes, then the square of its eccentricity.

Answer 1

$P\left(a \sec \frac{\pi }{6} , b \tan \frac{\pi }{6}\right)\equiv P\left(\frac{2 a}{\sqrt{3}} , \frac{b}{\sqrt{3}}\right)$
Therefore, equation of tangent at $P$ is $\frac{X}{\frac{\sqrt{3} a}{2}}-\frac{y}{\sqrt{3} b}=1$
$\therefore $ Area of the triangle $=\frac{1}{2}\times \frac{\sqrt{3} a}{2}\times \sqrt{3}b=3a^{2}$
$\Rightarrow \frac{b}{a}=4\therefore e^{2}=1+\frac{b^{2}}{a^{2}}=17$