Question

The length of the latus-rectum of the ellipse $3x^2+y^2=12$ is _______.

Answer 1

Given: Equation of ellipse is $3x^2+y^2=12$

$\Rightarrow \frac{x^2}{4}+\frac{y^2}{12}=1$

On comparing with standard form: $\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

We get, $a^2=4and{b}^2=12$

$\text{Length of latus - rectum}=\frac{2a^2}{b}=\frac{2\times 4}{\sqrt{12}}$

$\therefore \text{Length of latus- rectum}=\frac{4}{\sqrt{3}}$