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Question Number 41139 by ajfour last updated on 02/Aug/18

Commented by ajfour last updated on 02/Aug/18

$$Q.41117alternatesolution$$

Answered by ajfour last updated on 02/Aug/18

$$AP=2r\sin \theta$$$$PF=2r\sin \theta \cos \theta$$$$AF=2r\sin {}^2\theta$$$$Area(△APF)=A\left(\theta \right)=\frac{1}{2}\times PF\times AF$$$$A\left(\theta \right)=2r^2\sin {}^3\theta \cos \theta$$$$\frac{dA}{d\theta }=2r^2(3\sin {}^2\theta \cos {}^2\theta -\sin {}^4\theta )=0$$$$\Rightarrow \tan \theta =\sqrt{3}or\theta =60°$$$$therefore{A}_{max}=2r^2{\left(\frac{\sqrt{3}}{2}\right)}^3\left(\frac{1}{2}\right)$$$${\mathit{A}}_{\mathit{m}\mathit{a}\mathit{x}}=\frac{3\sqrt{3}}{8}{\mathit{r}}^2.$$

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