Question:
If a directrix of a hyperbola centred at the
origin and passing through the point $(4,-2 \sqrt{3})$ is $5 x=4 \sqrt{5}$ and its eccentricity is e, then :
Correct Option: 1
Solution:
Hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
$\frac{\mathrm{a}}{\mathrm{e}}=\frac{4}{\sqrt{5}}$ and $\frac{16}{\mathrm{a}^{2}}-\frac{12}{\mathrm{~b}^{2}}=1$
$\mathrm{a}^{2}=\frac{16}{5} \mathrm{e}^{2} \ldots(1)$ and $\frac{16}{\mathrm{a}^{2}}-\frac{12}{\mathrm{a}^{2}\left(\mathrm{e}^{2}-1\right)}=1$ .....(2)
From (1) & (2)
$16\left(\frac{5}{16 \mathrm{e}^{2}}\right)-\frac{12}{\left(\mathrm{e}^{2}-1\right)}\left(\frac{5}{16 \mathrm{e}^{2}}\right)=1$
$\Rightarrow 4 \mathrm{e}^{4}-24 \mathrm{e}^{2}+35=0$