Question

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

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

Answer 1

The correct option is B $\frac{4}{\sqrt{3}}$

Let $A^2=a^2+1$, $B^2=a^2+2$ {Here, $B^2>A^2$}

So, $e=\sqrt{1-\frac{A^2}{B^2}}=\sqrt{1-\frac{a^2+1}{a^2+2}}=\frac{1}{\sqrt{a^2+2}}=\frac{1}{\sqrt{3}}$

So, $\sqrt{a^2+2}=\sqrt{3}\Rightarrow a=\pm 1$

So, length of lotus return $=2\frac{A^2}{B}$

Length $=\frac{2(a^2+1)}{\sqrt{a^2+2}}=\frac{2\left(2\right)}{\sqrt{3}}=\frac{4}{\sqrt{3}}$