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

$\begin{array}{1 1}4\\3\\8\\\large\frac{4}{\sqrt 3}\end{array}$

• Length of the latus rectum of an ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1 when a < b is \large\frac{2a^2}{b} Answer : \large\frac{4}{\sqrt 3} Equation of the ellipse is 3x^2+y^2=12 (ie) \large\frac{x^2}{4}+\frac{y^2}{12}$$=1$
$\Rightarrow a^2=4$ or $a=2$
$b^2=12$ or $b=\sqrt{12}$
Length of the latus rectum =$\large\frac{2a^2}{b}$
$\Rightarrow \large\frac{2\times 4}{\sqrt{12}}$
$\Rightarrow \large\frac{4}{\sqrt 3}$
The length of the latus rectum of the ellipse $3x^2+y^2=12$ is $\large\frac{4}{\sqrt 3}$