The eccentricity of the hyperbola with length of the latus rectum as 8 and the length of its conjugate axis is equal to half of the distance between its foci , is

solution:-

In hyperbola length of latus rectum$=\frac{2b^2}{a}$

$\Rightarrow \frac{2b^2}{a}=8$ ……….$\left(1\right)$

Length of conjugate axis$=\frac{1}{2}$ (distance between the foci)

$\Rightarrow 2b=\frac{1}{2}\left(2ae\right)$

$\Rightarrow b=\frac{ae}{2}$

Now substituting in $\left(1\right)$ we get

$\frac{2}{a}\times \frac{a^2{e}^2}{4}=8$

$\Rightarrow a{e}^2=16$ ……….$\left(3\right)$

But $\frac{2b^2}{a}=8$

$\Rightarrow b^2=4a$

$\Rightarrow a(e^2-1)=4$ ……..$\left(4\right)$

Now $16-a=4$

$\Rightarrow a=12$ from $3$,

$12e^2=16$

$\Rightarrow e=\frac{2}{\sqrt{3}}$.