Question

Find the equation of hyperbola in standard form if the length of transverses axis is $6$ and its vertex bisects the distance between center and focus.

Answer 1

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$A{A}^{{}^{\prime }}=2a=6$

$a=3$

$\therefore a=\frac{ae}{2}$ (because it bisects).

$\Rightarrow 2a=ae$

$\Rightarrow e=2$

$\sqrt{1+\frac{b^2}{a^2}}=2$

$\Rightarrow 1+\frac{b^2}{a^2}=4$

$\Rightarrow 1+\frac{b^2}{a}=4$

$\Rightarrow 9+b^2=36$

$\Rightarrow b^2=27$

$\Rightarrow b=3\sqrt{3}$

Equation of Hyperbola:

$\frac{x^2}{9}-\frac{y^2}{27}=1$