Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The eccentricity of the hyperbola whose latus rectum is $8$ and conjugate axis is equal to half of the distance between the foci is

Conic Sections

Solution:

Let equation of hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Given, $\frac{2b^{2}}{a}=8$
$\Rightarrow \frac{b^{2}}{a}=4$
and $2b=\frac{1}{2}\left(2ae\right)$
$\Rightarrow 2b = ae$
$\Rightarrow 4b^{2} = a^{2}e^{2}$
$\Rightarrow \, 4\left(\frac{b^{2}}{a^{2}}\right)=e^{2}$
$\Rightarrow 4\left(e^{2}-1\right)=e^{2} \left[\because\, b^{2}=a^{2} \left(e^{2}-1\right)\right]$
$\Rightarrow 3e^{2} = 4$
$\Rightarrow e=\frac{2}{\sqrt{3}}$