$Let a and b be positive real numbers such that b < a; Let P (2, y_{1}) be a point in the$
$first quadrant that lies on ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . Suppose the tangent to the ellipse at P$
$passes through (8, 0) and suppose the normal to ellipse at P has slope \frac{4}{\sqrt{3}}, then$
$Area of triangle formed by tangent at P, the normal at P and x -axis is$

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$\frac{171}{16} \sqrt{3} sq. units$

$\frac{171}{32} \sqrt{3} sq. units$

$\frac{171}{2} \sqrt{2} sq.units$

$\frac{171}{8} \sqrt{3} sq. units$

answer is D.

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Detailed Solution

$\frac{2 x}{a^{2}} + \frac{y}{b^{2}} = 1 - - - - - (1)$

$\begin{array}{l} m_{r} = \frac{- \sqrt{3}}{4} \\ y - 0 = - \frac{\sqrt{3}}{4} (x - 8) \\ 4 y = - \sqrt{3} x + 8 \sqrt{3} \\ \sqrt{3} x + 4 y = 8 \sqrt{3} \\ C o m p a r i n g t h e \tan g e n t e q u a t i o n s \\ \frac{2}{a^{2} \sqrt{3}} = \frac{y_{1}}{b^{2} 4} = \frac{1}{8 \sqrt{3}} \Rightarrow 2 = \frac{a^{2}}{8}; \frac{b^{2}}{2 \sqrt{3}} = y \\ a^{2} = 16 \end{array}$