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

Answer: 17

$P\left(a\sec \frac{\pi }{6},b\tan \frac{\pi }{6}\right)\equiv P\left(\frac{2a}{\sqrt{3}},\frac{b}{\sqrt{3}}\right)$
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 $=\frac{1}{2}\times \frac{\sqrt{3}a}{2}\times \sqrt{3}b=3a^2$
$\Rightarrow \frac{b}{a}=4\therefore e^2=1+\frac{b^2}{a^2}=17$