Prove that the greatest area which the $\Delta APY$ can have is $3\sqrt3\frac{a^2}{8}$sq units

Question

From a fixed point $A$ on the circumference of a circle of radius $a$,let the perpendicular $AY$ fall on the tangent at a point $P$ on the circle,prove that the greatest area which the $\Delta APY$ can have is $3\sqrt3\frac{a^2}{8}$sq units.

I cannot solve this question.How to determine the area of $\Delta APY$ which is to be maximized by differentiating.Please help me.

A.Γ. · Answer 1 · 2015-08-26T08:56:03.503

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Hint: place the center of the circle at the origin and the point $A$ at $(-a,0)$. An arbitrary point $P$ on the circle can be described as $(a\cos t, a\sin t)$, $t\in[0,2\pi]$. Express the area of the corresponding right angle triangle $APY$ as a function of $t$ and maximize it.

edited Aug 26 '15 at 08:56

answered Aug 26 '15 at 07:56

A.Γ.

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Sir,$Y$ is not on the circle.Point P is on the circle. – Vinod Kumar Punia Aug 26 '15 at 08:44
@VinodKumarPunia OK, sorry, just exchange $P\leftrightarrow Y$. – A.Γ. Aug 26 '15 at 08:56
The area of $APY$ calculated using this method comes out to be $\frac{1}{2}a^2\cot^2\frac{t}{2}$ which does not maximize to desired value,Sir. – Vinod Kumar Punia Sep 02 '15 at 15:29
@VinodKumarPunia It does not look right. If you set $t=0$ you'll get $\infty$. How is it possible for the area to be so large? – A.Γ. Sep 02 '15 at 16:16

Prove that the greatest area which the $\Delta APY$ can have is $3\sqrt3\frac{a^2}{8}$sq units

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