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If the line x - 2y = 12 is tangent to the ellipse

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 :

  1. 9

  2. $8 \sqrt{3}$

  3. $12 \sqrt{2}$

  4. 5

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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