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. $4e^4 - 12e^2 -27 = 0$
• D. $4e^4 -24e^2 + 27 = 0 $

Answer 1

Answer: (A)

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