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

Answer 1

Answer: (A)

$4e^4-24e^2+35=0$

Hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\frac{a}{e}=\frac{4}{\sqrt{5}}$ and $\frac{16}{a^2}-\frac{12}{b^2}=1$
$a^2=\frac{16}{5}{e}^2$ ...(1)
and $\frac{16}{a^2}-\frac{12}{a^2(e^2-1)}=1$ ...(2)
For (1) & (2)
$16\left(\frac{5}{16e^2}\right)-\frac{12}{(e^2-1)}\left(\frac{5}{16e^2}\right)=1$
$\Rightarrow 4e^4-24e^2+35=0$