The area of the quadrilateral formed by the tangents at end points of latus rectum to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ is __________

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$\frac{27}{4}$

$\frac{27}{2}$

$27$

answer is D.

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Detailed Solution

$a = 3, b = \sqrt{5} \Rightarrow e = \frac{2}{3}$
Area of triangle formed by tangents with the co-ordinate axes $= \frac{1}{2} (\frac{9}{2}) (3) = \frac{27}{4}$
$∴$ Required area $= 4 (\frac{27}{4}) = 27$

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