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Question Number 47559 by ajfour last updated on 11/Nov/18

Commented by ajfour last updated on 11/Nov/18

$$Q.47519\left(Solutionattempted\right)$$

Answered by ajfour last updated on 11/Nov/18

$$Area△ABC=\frac{1}{2}\left(AB\right)\left(CP\right)$$$$AB=\sqrt{x_A^2+y_B^2}$$$$Let\mathrm{\angle }OAB=\theta ;CP=l$$$$P(x_1,y_1),C(x_C,y_C)$$$$Eq.oftangenttoEllipseatP$$$$\frac{{xx}_1}{a^2}+\frac{{yy}_1}{b^2}=1$$$$\tan \theta =\frac{b^2{x}_1}{a^2{y}_1}\Rightarrow \frac{x_1}{y_1}=\frac{a^2\sin \theta }{b^2\cos \theta }..\left(i\right)$$$$\Rightarrow x_A=\frac{a^2}{x_1};y_B=\frac{b^2}{y_1}$$$$x_C=x_1-l\sin \theta$$$$y_C=y_1-l\cos \theta$$$$AndsinceCliesonellipse$$$$\frac{{(x_1-l\sin \theta )}^2}{a^2}+\frac{{(y_1-l\cos \theta )}^2}{b^2}=1$$$$\Rightarrow \frac{l^2\sin {}^2\theta }{a^2}+\frac{l^2\cos {}^2\theta }{b^2}+(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2})$$$$-2l(\frac{x_1\sin \theta }{a^2}+\frac{y_1\cos \theta }{b^2})=1$$$$but(\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2})=1,So$$$$\Rightarrow l=\frac{2(\frac{x_1\sin \theta }{a^2}+\frac{y_1\cos \theta }{b^2})}{\frac{\sin {}^2\theta }{a^2}+\frac{\cos {}^2\theta }{b^2}}$$$$△=\frac{l}{2}\sqrt{x_A^2+y_A^2}$$$$=\frac{(\frac{x_1\sin \theta }{a^2}+\frac{y_1\cos \theta }{b^2})\sqrt{\frac{a^4}{x_1^2}+\frac{b^4}{y_1^2}}}{\frac{\sin {}^2\theta }{a^2}+\frac{\cos {}^2\theta }{b^2}}$$$$\Rightarrow △=\frac{(\frac{x_1\sin \theta }{y_1{a}^2}+\frac{\cos \theta }{b^2})\sqrt{\frac{y_1^2{a}^4}{x_1^2}+b^4}}{\frac{\sin {}^2\theta }{a^2}+\frac{\cos {}^2\theta }{b^2}}$$$$using\left(i\right):\frac{x_1}{y_1}=\frac{a^2\sin \theta }{b^2\cos \theta }$$$$△=\frac{(\frac{\sin {}^2\theta }{b^2\cos \theta }+\frac{\cos \theta }{b^2})\sqrt{\frac{b^4\cos {}^2\theta }{\sin {}^2\theta }+b^4}}{\frac{\sin {}^2\theta }{a^2}+\frac{\cos {}^2\theta }{b^2}}$$$$△=\frac{\frac{1}{b^2\cos \theta }\times \frac{b^2}{\sin \theta }}{\frac{\sin {}^2\theta }{a^2}+\frac{\cos {}^2\theta }{b^2}}$$$$\Rightarrow \frac{1}{△}=\sin \theta \cos \theta (\frac{\sin {}^2\theta }{a^2}+\frac{\cos {}^2\theta }{b^2})$$$$let\tan \theta =m$$$$\Rightarrow \frac{1}{△}=\frac{m(\frac{m^2}{a^2}+\frac{1}{b^2})}{{(1+m^2)}^2}$$$$\frac{d(1/△)}{d\theta }=\frac{d(1/△)}{dm}\times \frac{dm}{d\theta }$$$$\frac{d(1/△)}{d\theta }=0(forminimum△)$$$$\Rightarrow \frac{d(1/△)}{dm}={(1+m^2)}^2(\frac{3m^2}{a^2}+\frac{1}{b^2})$$$$-4m^2(1+m^2)(\frac{m^2}{a^2}+\frac{1}{b^2})=0$$$$\Rightarrow (1+m^2)(\frac{3m^2}{a^2}+\frac{1}{b^2})=4m^2(\frac{m^2}{a^2}+\frac{1}{b^2})$$$$\frac{3m^2}{a^2}+\frac{1}{b^2}+\frac{3m^4}{a^2}+\frac{m^2}{b^2}=\frac{4m^4}{a^2}+\frac{4m^2}{b^2}$$$$\Rightarrow m^4+3m^2(\frac{a^2}{b^2}-1)-\frac{a^2}{b^2}=0$$$$m^2=\sqrt{\frac{9}{4}{(\frac{a^2}{b^2}-1)}^2+\frac{a^2}{b^2}}-\frac{3}{2}(\frac{a^2}{b^2}-1)$$$${△}_{min}=\frac{a^2{(1+m^2)}^2}{m(m^2+\frac{a^2}{b^2})}.$$$$\left(Asaspecialcase\right)$$$$Ifa=b=R,thenm=1$$$${△}_{min}=2R^2.$$

Commented by mr W last updated on 11/Nov/18

$$greatsolutionsir!$$

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