Question

For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ distance between the foci is $10.$ From the point $(2,\sqrt{3}),$ perpendicular tangents are drawn to the hyperbola. If eccentricity of the hyperbola is $e,$ then the value of $4e$ is

Answer 1

Director circle of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2-b^2.$
Since, $P(2,\sqrt{3})$ is on director circle,
$\Rightarrow a^2-b^2=7\dots \left(1\right)$
and $F_1{F}_2=2ae=10$
$\Rightarrow 2a\sqrt{1+\frac{b^2}{a^2}}=10$
$\Rightarrow \sqrt{a^2+b^2}=5$
$\Rightarrow a^2+b^2=25\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$a=4,b=3$

$e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+{\left(\frac{3}{4}\right)}^2}=\frac{5}{4}$
$\Rightarrow 4e=5$