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