Let point P lie on the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1$ and normal to ellipse at P meets the coordinate axes in A and B. If O is the origin and M is the foot of perpendicular from origin to AB, then

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Maximum area of $Δ A O B$ is $\frac{81}{80}$ sq.units

maximum value of OM is 2 unit

maximum area of $Δ A O B$ is 2.025 sq.units

maximum value of OM is 1 unit

answer is C, D.

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Detailed Solution

Equation of normal is $5 x \sec α - 4 y \cos e c α = 9$ , coordinate of A thus $(\frac{9}{5} \cos α, 0)$ is and B is $(0, - \frac{9}{5} \sin α)$
Area of triangle AOB is $\frac{1}{2} . \frac{9}{5} \cos α . \frac{9}{4} \sin α = \frac{81}{80} \sin 2 α$
Thus maximum area is $\frac{81}{80}$ sq.units
$O M = \frac{9}{\sqrt{25 \sec^{2} α + 16 \cos e c^{2} α}}$
$\frac{9}{\sqrt{41 + (25 \tan^{2} α + 16 \cot^{2} α)}} \leq \frac{9}{\sqrt{41 + 2. \sqrt{25.16}}} = 1$