Question

If the eccentricity of the ellipse $\frac{x^2}{a^2+1}+\frac{y^2}{a^2+2}=1$ is $\frac{1}{\sqrt{6}}$, then latus rectum of ellipse is:

• A. $\frac{5}{\sqrt{6}}$
• B. $\frac{10}{\sqrt{6}}$
• C. $\frac{8}{\sqrt{6}}$
• D. None of these

Answer 1

Answer: (B)

$\frac{10}{\sqrt{6}}$

Here $a^2+2>a^2+1$
$\Rightarrow a^2+1=(a^2+2)(1-e^2)$
$\Rightarrow a^2+1=(a^2+2)\frac{5}{6}$
$\Rightarrow 6a^2+6=5a^2+10$
$\Rightarrow a^2=10-6=4$
$\Rightarrow a=2$
Latus rectum $=\frac{2(a^2+1)}{\sqrt{a^2+2}}=\frac{2\times 5}{\sqrt{6}}=\frac{10}{\sqrt{6}}$