let the line $y = m x$ and the ellipse $4 x^{2} + 9 y^{2} = 1$ intersect at point P in the first quadrant. If the normal to the ellipse at P meets the coordinate axes at $A (α, 0)$ and $B (0, - \frac{5}{8 \sqrt{3}})$ , then area of triangle OAB is equal to (where O is origin)

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$\frac{1}{16 \sqrt{3}}$

$\frac{25}{288 \sqrt{3}}$

$\frac{25}{576 \sqrt{3}}$

$\frac{9}{144 \sqrt{3}}$

answer is B.

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Detailed Solution

$a^{2} = \frac{1}{4}, b^{2} = \frac{1}{9} l e t P b e (x_{1}, y_{1})$
Equation normal $\frac{1 x}{4 x_{1}} - \frac{1 y}{9 y_{1}} = \frac{5}{36}$
It cuts y-axis at $(0, - \frac{5 y_{1}}{4}) \Rightarrow \frac{- 5 y_{1}}{4} = \frac{- 5}{8 \sqrt{3}} \Rightarrow y_{1} = \frac{1}{2 \sqrt{3}}$
$∴ x_{1} = \frac{1}{4}$
Thus $α = \frac{5}{36}$
$A r e a o f Δ O A B = \frac{1}{2} \times \frac{5}{36} \times \frac{5}{8 \sqrt{3}} = \frac{25}{576 \sqrt{3}}$