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Question Number 54242 by ajfour last updated on 01/Feb/19

Answered by mr W last updated on 01/Feb/19

$$parabola:$$$$y=h-\frac{x^2}{c}$$$$ellipse:$$$$\frac{x^2}{a^2}+\frac{{(y-b)}^2}{b^2}=1$$$$intersection:$$$$\frac{c(h-y)}{a^2}+\frac{{(y-b)}^2}{b^2}=1$$$${cb}^2(h-y)+a^2{(y-b)}^2=a^2{b}^2$$$$a^2{y}^2-(2a^2+cb)by+{chb}^2=0$$$$D={(2a^2+cb)}^2{b}^2-4a^2{chb}^2=0$$$${(2a^2+cb)}^2-4a^{2}ch=0$$$$\Rightarrow b=\frac{2a}{c}(\sqrt{ch}-a)$$$$A=\pi ab=\frac{2\pi }{c}{a}^2(\sqrt{ch}-a)$$$$\frac{dA}{da}=0$$$$\Rightarrow 2a\sqrt{ch}-3a^2=0$$$$\Rightarrow a=\frac{2\sqrt{ch}}{3}$$$$\Rightarrow b=\frac{2}{c}\times \frac{2\sqrt{ch}}{3}(\sqrt{ch}-\frac{2\sqrt{ch}}{3})=\frac{4h}{9}$$$$A_{max}=\pi ab=\frac{8\pi h\sqrt{ch}}{27}$$

Commented by ajfour last updated on 01/Feb/19

$$\mathcal{THANK}\mathcal{YOU}\mathbb{SIR}.$$

Answered by ajfour last updated on 01/Feb/19

$$letCbeorigin.$$$$eq.ofparabola:y=\mathit{h}-b-\frac{x^2}{c}$$$$letP(a\cos \theta ,b\sin \theta )$$$$\frac{dy}{dx}{\mid }_P=-\frac{b\cos \theta }{a\sin \theta }=-\frac{2x}{c}=-\frac{2a\cos \theta }{c}$$$$\Rightarrow a^2=\frac{bc}{2\sin \theta }....\left(i\right)$$$$b\sin \theta =\mathit{h}-b-\frac{a^2\cos {}^2\theta }{c}$$$$using\left(i\right)herein$$$$b\sin \theta =\mathit{h}-b-\frac{b\cos {}^2\theta }{2\sin \theta }$$$$\Rightarrow 2b\sin {}^2\theta =2\mathit{h}\sin \theta -2b\sin \theta -b\cos {}^2\theta$$$$\Rightarrow b(\sin {}^2\theta +2\sin \theta +1)=2\mathit{h}\sin \theta$$$$\Rightarrow b=\frac{2\mathit{h}\sin \theta }{{(1+\sin \theta )}^2}andusing\left(i\right)$$$$\Rightarrow a^2=\frac{c\mathit{h}}{{(1+\sin \theta )}^2}$$$$\Rightarrow a=\frac{\sqrt{c\mathit{h}}}{1+\sin \theta }$$$$\Rightarrow A=\pi ab=\frac{2\pi h\sqrt{ch}\sin \theta }{{(1+\sin \theta )}^3}.....\left(ii\right)$$$$letS=\frac{\sin \theta }{{(1+\sin \theta )}^3}$$$$\frac{dS}{d\theta }=0\Rightarrow$$$$\cos \theta {(1+\sin \theta )}^3=3\sin \theta \cos \theta {(1+\sin \theta )}^2$$$$\Rightarrow 1+\sin \theta =3\sin \theta$$$$or\sin \theta =1/2\Rightarrow S=\frac{4}{27}$$$$\Rightarrow A_{max}=\frac{8\pi \mathit{h}\sqrt{c\mathit{h}}}{27}.$$$$(parametric{isn}^{\prime }twrongeither!)$$

Commented by mr W last updated on 01/Feb/19

$$nicetoo!$$

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