Question

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

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

Answer 1

Answer: (C)

$\because \left(3,-\frac{9}{2}\right)$ lies on $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\Rightarrow \frac{9}{a^2}+\frac{81}{4b^2}=1$ …… $\left(1\right)$ Equation of the tangent at $\left(3,-\frac{9}{2}\right)$ is $\frac{3x}{a^2}+\frac{-\frac{9}{2}y}{b^2}=1$
& given equation of the tangent is:
$x-2y=12\Rightarrow \frac{x}{12}+\frac{-y}{6}=1$
On comparing these equations:
$\frac{a^2}{3}=12\Rightarrow a^2=36\Rightarrow a=6$
$\frac{2b^2}{9}=6\Rightarrow b^2=27\Rightarrow b=3\sqrt{3}$
Therefore, the length of latus rectum
$=\frac{2b^2}{a}=\frac{2\times 27}{6}=9$