If lines x -2y = 12 is tangent to ellipse frac{{{x^2}}}{{{a^2}}} + frac{{{y^2}}}{{{b^2}}} = 1 at poi

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10-2. Parabola, Ellipse, Hyperbola

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If the lines $x -2y = 12$ 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

A

$12\sqrt 2$

B

$9$

C

$8\sqrt 3$

D

$5$

(JEE MAIN-2019)

Solution

Tangent at $\left( {3,\frac{{ – 9}}{2}} \right)$

$\frac{{3x}}{{{a^2}}} – \frac{{9y}}{{2{b^2}}} = 1$

Comparing with $x – 2y = 12$

$\frac{3}{{{a^2}}} – \frac{9}{{4{b^2}}} = \frac{1}{{12}}$

$ \Rightarrow a = 6$ and $b = 3\sqrt 3 $

Length of latus rectum $ = \frac{{2{b^2}}}{a} = 9$

Standard 11

Mathematics

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