Question

Eccentricity of ellipse $\frac{x^2}{a^2+1}+\frac{y^2}{a^2+2}=1$ is $\frac{1}{\sqrt{3}}$ then length of Latusrectum is

• A. $\frac{8}{\sqrt{3}}$
• B. $\frac{4}{\sqrt{3}}$
• C. $2\sqrt{3}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A $\frac{8}{\sqrt{3}}$

$\frac{x^2}{a^2+1}+\frac{y^2}{a^2+2}=1$

Eccentricity of ellipse $=\frac{1}{\sqrt{3}}$

So, $\frac{\sqrt{b^2-a^2}}{a}=\frac{1}{\sqrt{3}}$

Where ellipse

$\frac{x^2}{a^2+1}+\frac{y^2}{a^2+2}=1$

So, here

$\frac{\sqrt{a^2+2a^2-1}}{a}=\frac{1}{\sqrt{3}}$

$⟹a^2=2$

So, equation is

$\frac{x^2}{3}+\frac{y^2}{4}=1$

Latus rectum $=\frac{2b^2}{a}=\frac{2\times 4}{\sqrt{3}}=\frac{8}{\sqrt{3}}$