Question

If a directrix of a hyperbola centred at the origin and passing through the point $(4,-2\sqrt{3})$ is $5x=4\sqrt{5}$ and its eccentricity is $e$, then :

• A. $4e^4-24e^2+27=0$
• B. $4e^4-12e^2-27=0$
• C. $4e^4+8e^2-35=0$
• D. $4e^4-24e^2+35=0$

Answer 1

The correct option is D $4e^4-24e^2+35=0$

Since, the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passes through $(4,-2\sqrt{3})$
Therefore, $\frac{16}{a^2}-\frac{12}{b^2}=1$
$\Rightarrow 16-12\frac{a^2}{b^2}=a^2\cdots \left(1\right)$

Directrix : $5x=4\sqrt{5}$
$\Rightarrow \frac{4}{\sqrt{5}}=\frac{a}{e}\Rightarrow 16e^2=5a^2\cdots \left(2\right)$

From eqn $\left(1\right)$ and $\left(2\right)$
$\Rightarrow 16-\frac{12}{e^2-1}=\frac{16e^2}{5}\Rightarrow 80(e^2-1)-60=16e^2(e^2-1)\Rightarrow 4e^4-24e^2+35=0$