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10-2. Parabola, Ellipse, Hyperbola
hard
If a directrix of a hyperbola centered at the origin and passing through the point $(4, -2\sqrt 3)$ is $5x = 4\sqrt 5$ and its eccentricity is $e$, then
A
$4e^4 + 8e^2 -35 = 0$
B
$4e^4 -24e^2 + 35 = 0$
C
$4e^4 -12e^2 -27 = 0$
D
$4e^4 -24e^2 + 27 = 0$
(JEE MAIN-2019)
Solution
Let hyperbola be $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ and passes through $\left( {4, – 2\sqrt 3 } \right)$ therefore
$\frac{{16}}{{{a^2}}} – \frac{{12}}{{{b^2}}} = 1\,\,\,\,\,\,\,\,\,\,\,……\left( i \right)$
$\because$ ${b^2} = {a^2}\left( {{e^2} – 1} \right)$
$x = \frac{{4\sqrt 5 }}{5} = \frac{e}{a} \Rightarrow {a^2} = \frac{{16}}{5}{e^2}\,\,\,\,\,\,\,\,\,\,………\left( {ii} \right)$
On solving $(i)$ and $(ii)$
$ \Rightarrow 4{e^2}\, – 24{e^2}\, + 35 = 0$
Standard 11
Mathematics
