If a directrix of a hyperbola centred at the

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 :

  1. $4 e^{4}-24 e^{2}+35=0$

  2. $4 \mathrm{e}^{4}+8 \mathrm{e}^{2}-35=0$

  3. $4 e^{4}-12 e^{2}-27=0$

  4. $4 \mathrm{e}^{4}-24 \mathrm{e}^{2}+27=0$


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)

$16\left(\frac{5}{16 \mathrm{e}^{2}}\right)-\frac{12}{\left(\mathrm{e}^{2}-1\right)}\left(\frac{5}{16 \mathrm{e}^{2}}\right)=1$

$\Rightarrow 4 \mathrm{e}^{4}-24 \mathrm{e}^{2}+35=0$ 

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