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Question Number 66656 by Tanmay chaudhury last updated on 18/Aug/19

Commented by Rahul Kumar last updated on 18/Aug/19

$$pleaseanswerthequestion.$$

Answered by mr W last updated on 18/Aug/19

Commented by mr W last updated on 18/Aug/19

$$letr=radiusofthecircles$$$$let\theta =\frac{\mathrm{\angle }AOB}{2}$$$$eqn.ofellipse:$$$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$withb=r+2r\cos \theta =r(1+2\cos \theta )$$$$x_B=2r\sin \theta$$$$eqn.ofcircleatB:$$$${(x-x_B)}^2+y^2=r^2$$$$$$$$intersectionofcircleBandellipse:$$$$\frac{x^2}{a^2}+\frac{r^2-{(x-x_B)}^2}{b^2}=1$$$$\frac{x^2}{a^2}+\frac{r^2-x^2-x_B^2+2x_{B}x}{b^2}=1$$$$\frac{b^2}{a^2}{x}^2+r^2-x^2-4r^2{\sin }^2\theta +4r\sin \theta x=b^2$$$$(1-\frac{b^2}{a^2})x^2-4r\sin \theta x-(r^2-4r^2{\sin }^2\theta -b^2)=0$$$$duetotangency$$$$\mathrm{\Delta }=16r^2{\sin }^2\theta +4(1-\frac{b^2}{a^2})(r^2-4r^2{\sin }^2\theta -b^2)=0$$$$\Rightarrow 4r^2{\sin }^2\theta +r^2(1-\frac{b^2}{a^2})(1-4{\sin }^2\theta -\frac{b^2}{r^2})=0$$$$\Rightarrow {4\sin }^2\theta +(1-\frac{b^2}{a^2})[1-4{\sin }^2\theta -{(1+2\cos \theta )}^2]=0$$$$\Rightarrow {\sin }^2\theta -(1-\frac{b^2}{a^2})(1+\cos \theta )=0$$$$\Rightarrow 1-\cos \theta =1-\frac{b^2}{a^2}$$$$\Rightarrow \frac{b^2}{a^2}=\cos \theta$$$$\Rightarrow a^2=\frac{b^2}{\cos \theta }$$$$letP=\frac{a^2{b}^2}{r^4}=\frac{b^4}{r^4\cos \theta }=\frac{{(1+2\cos \theta )}^4}{\cos \theta }$$$$with\lambda =\cos \theta$$$$\Rightarrow P=\frac{{(1+2\lambda )}^4}{\lambda }$$$$Areaofellipse=\pi ab$$$$minimumareaofellipsemeansalso$$$$minimumvalueofP,therefore$$$$\frac{dP}{d\lambda }=\frac{8{(1+2\lambda )}^3}{\lambda }-\frac{{(1+2\lambda )}^4}{{\lambda }^2}=0$$$$8-\frac{1+2\lambda }{\lambda }=0$$$$\Rightarrow 6\lambda =1$$$$\Rightarrow \lambda =\cos \theta =\frac{1}{6}$$$$\Rightarrow \sin \theta =\frac{\sqrt{35}}{6}$$$$\Rightarrow \tan \theta =\sqrt{35}$$$$\Rightarrow \mathrm{\angle }AOB=2\theta =2{\tan }^{-1}\sqrt{35}$$

Commented by mr W last updated on 18/Aug/19

$$thankssir!$$$$canyoupleasehelpcheckingmy$$$$solution.myresultdiffersfrom$$$$theoptionsgiveninbook.$$

Commented by Tanmay chaudhury last updated on 18/Aug/19

$$siryouareunique...$$

Commented by MJS last updated on 18/Aug/19

$$\mathrm{I}\mathrm{get}\mathrm{the}\mathrm{same}\mathrm{result}$$

Commented by mr W last updated on 18/Aug/19

$$thanksforcomfirmingsir!$$$$thatmeanstheanswergiveninbook$$$$iswrong.$$

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