Question:
If the line $\mathrm{x}-2 \mathrm{y}=12$ 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 recturm of the ellipse is :
Correct Option: 1
Solution:
Tangent at $\left(3,-\frac{9}{2}\right)$
$\frac{3 x}{a^{2}}-\frac{9 y}{2 b^{2}}=1$
Comparing this with $x-2 y=12$
$\frac{3}{a^{2}}=\frac{9}{4 b^{2}}=\frac{1}{12}$
we get $a=6$ and $b=3 \sqrt{3}$
$\mathrm{L}(\mathrm{LR})=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=9$
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