The eccentricity of an ellipse E with centre at the origin O is √32 and its directrices ar…

Question

The eccentricity of an ellipse $E$ with centre at the origin $O$ is $\frac{\sqrt{3}}{2}$ and its directrices are $x=\pm \frac{4\sqrt{6}}{3}$. Let $H:\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$ be a hyperbola whose eccentricity is equal to the length of semi-major axis of E , and whose length of latus rectum is equal to the length of minor axis of E . Then the distance between the foci of H is :

Accepted Answer

For the ellipse $E$, let its equation be

$\frac{{x}^{2}}{{A}^{2}}+\frac{{y}^{2}}{{B}^{2}}=1,\, A>B$

since its major axis is along the $x$-axis because the directrices are of the form $x=$.

Given eccentricity,

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

For an ellipse with major axis along the $x$-axis, the directrices are

$x=\pm \frac{A}{e}$

We are given

$x=\pm \frac{4\sqrt{6}}{3}$

So,

$\frac{A}{e}=\frac{4\sqrt{6}}{3}$

Substituting $e=\frac{\sqrt{3}}{2}$,

$A=\frac{\sqrt{3}}{2}\cdot \frac{4\sqrt{6}}{3}$

$A=\frac{4\sqrt{18}}{6}=\frac{4\cdot 3\sqrt{2}}{6}=2\sqrt{2}$

Thus the semi-major axis of ellipse $E$ is

$A=2\sqrt{2}$

Now for the ellipse,

${B}^{2}={A}^{2}(1-{e}^{2})$

Since

${A}^{2}=(2\sqrt{2}{)}^{2}=8$

and

${e}^{2}=\frac{3}{4}$

we get

${B}^{2}=8(1-\frac{3}{4})=8\cdot \frac{1}{4}=2$

Hence

$B=\sqrt{2}$

So the length of minor axis of $E$ is

$2B=2\sqrt{2}$

Now consider the hyperbola

$H:\frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1$

Its eccentricity is equal to the length of semi-major axis of $E$, so

${e}_{H}=2\sqrt{2}$

For a hyperbola,

${e}_{2}^{H}=1+\frac{{b}^{2}}{{a}^{2}}$

Thus,

$(2\sqrt{2}{)}^{2}=1+\frac{{b}^{2}}{{a}^{2}}$

$8=1+\frac{{b}^{2}}{{a}^{2}}$

$\frac{{b}^{2}}{{a}^{2}}=7\, \Rightarrow \, {b}^{2}=7{a}^{2}$

Also, the length of latus rectum of the hyperbola is equal to the length of minor axis of $E$.

For the hyperbola, length of latus rectum is

$\frac{2{b}^{2}}{a}$

So,

$\frac{2{b}^{2}}{a}=2\sqrt{2}$

$\frac{{b}^{2}}{a}=\sqrt{2}$

Using ${b}^{2}=7{a}^{2}$,

$\frac{7{a}^{2}}{a}=\sqrt{2}$

$7a=\sqrt{2}$

$a=\frac{\sqrt{2}}{7}$

Now for the hyperbola,

$c=a{e}_{H}$

So,

$c=\frac{\sqrt{2}}{7}\cdot 2\sqrt{2}=\frac{4}{7}$

Distance between the foci $=2c$,

$2c=\frac{8}{7}$

Hence, the correct answer is

$\frac{8}{7}$

So, Option D is correct.

Solutions · 1