Question

If the line $x-2y=12$ is tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $(3,-\frac{9}{2})$, then the length of the latus rectum of ellipse is :

• A. $9$
• B. $8\sqrt{3}$
• C. $5$
• D. $12\sqrt{2}$

Answer 1

The correct option is A $9$

$\frac{x}{12}-\frac{y}{6}=1\cdots \left(1\right)$ is tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $(3,-\frac{9}{2})$

Equation of tangent to the ellipse at the point $(3,-\frac{9}{2})$ is given by,
$\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$
$\Rightarrow \frac{3x}{a^2}-\frac{9y}{2b^2}=1\cdots \left(2\right)$
On equating $\left(1\right)$ and $2$, we have
$\frac{a^2}{3}=12\Rightarrow a^2=36$
$\frac{2b^2}{9}=6\Rightarrow 2b^2=54$

Therefore, length of latus rectum $=\left|\frac{2b^2}{a}\right|=9$