If a directrix of a hyperbola centred at the origin and passing through the point $(4,-2 \sqrt{3})$ is $\sqrt{5} x=4$ and $e$ is its eccentricity, then $e^2=$

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Aicharya · Accepted Answer

Since hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passes through $(4,-2 \sqrt{3})$ $	herefore \frac{16}{a^2}-\frac{12}{b^2}=1 \Rightarrow 16-12 \frac{a^2}{b^2}=a^2$ $\qquad ...\mathrm{(i)}$ Directrix : $\sqrt{5} x=4 \Rightarrow \frac{4}{\sqrt{5}}=\frac{a}{e}$ $\Rightarrow 16 e^2=5 a^2$ $\qquad ...\mathrm{(ii)}$ From eq. (i) and (ii), $\begin{aligned} & \Rightarrow 16-\frac{12}{e^2-1}=\frac{16 e^2}{5} \Rightarrow 4 e^4-24 e^2+35=0 \ & \Rightarrow e^2=\frac{7}{2}, \frac{5}{2} \end{aligned}$

If a directrix of a hyperbola centred at the origin and passing through the point $(4,-2 \sqrt{3})$ is…

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