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

 

  1. (1) 9

  2. (2) $12 \sqrt{2}$

  3. (3) 5

  4. (4) $8 \sqrt{3}$


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 $=\frac{2 b^{2}}{a}=\frac{2 \times 27}{6}=9$

 

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