Question:
If the line $x-2 y=12$ is 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 :
Correct Option: 1
Solution:
Equation of tangent to $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $\left(3,-\frac{9}{2}\right)$ is,
$\frac{3 x}{a^{2}}-\frac{9 y}{2 b^{2}}=1$
But given equation of tangent is, $x-2 y=12$
$\therefore \frac{3}{a^{2}}=\frac{-9}{2 b^{2} \cdot(-2)}=\frac{1}{12}$ (On comparing)
$\Rightarrow a^{2}=3 \times 12$ and $b^{2}=\frac{9 \times 12}{4}$
$\Rightarrow a=6$ and $b=3 \sqrt{3}$
Therefore, latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 27}{6}=9$