If a directrix of a hyperbola centered at the | vedclass

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Question

Let hyperbola be $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ and passes through $\left( {4, - 2\sqrt 3 } \right)$ therefore

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Vedclass · Accepted Answer

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

Vedclass · Answer

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$

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

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