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Question Number 60527 by ajfour last updated on 21/May/19

Commented by ajfour last updated on 21/May/19

$$\mathrm{Find}\mathrm{maximum}\mathrm{area}\mathrm{of}\mathrm{quadrilateral}$$$$\mathrm{OAPB}.\mathrm{The}\mathrm{ellipse}\mathrm{equation}\mathrm{is}\mathrm{the}$$$$\mathrm{usual}\mathrm{one}.$$

Answered by ajfour last updated on 22/May/19

$$\mathrm{Area}\mathrm{A}=\frac{1}{2}\mathrm{absin}\theta +\frac{1}{2}\mathrm{abcos}\theta$$$$\frac{\mathrm{dA}}{\mathrm{d}\theta }=0\Rightarrow \tan \theta =1$$$${\mathrm{A}}_{\max }=\frac{\mathrm{ab}}{\sqrt{2}}.$$

Commented by mr W last updated on 22/May/19

$$thankyousir!$$

Commented by ajfour last updated on 22/May/19

$$\mathrm{yes}\mathrm{sir},\mathrm{mistake};\mathrm{P}(\mathrm{acos}\theta ,\mathrm{bsin}\theta ).$$

Answered by mr W last updated on 22/May/19

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$\frac{2x}{a^2}+\frac{2{yy}^{\prime }}{b^2}=0$$$$tangentatP//AB:$$$$\frac{2a\cos \theta }{a^2}+\frac{2b\sin \theta }{b^2}(-\frac{b}{a})=0$$$$\cos \theta -\sin \theta =0$$$$\Rightarrow \theta =45°forpointP.$$$$A_{max}=\frac{ab\cos \theta +ba\sin \theta }{2}=\frac{\sqrt{2}ab}{2}$$

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