Question Number 66656 by Tanmay chaudhury last updated on 18/Aug/19
Commented by Rahul Kumar last updated on 18/Aug/19
$${please}\:{answer}\:{the}\:{question}. \\ $$
Answered by mr W last updated on 18/Aug/19
Commented by mr W last updated on 18/Aug/19
$${let}\:{r}={radius}\:{of}\:{the}\:{circles} \\ $$$${let}\:\theta=\frac{\angle{AOB}}{\mathrm{2}} \\ $$$${eqn}.\:{of}\:{ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${with}\:{b}={r}+\mathrm{2}{r}\:\mathrm{cos}\:\theta={r}\left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\theta\right) \\ $$$${x}_{{B}} =\mathrm{2}{r}\:\mathrm{sin}\:\theta \\ $$$${eqn}.\:{of}\:{circle}\:{at}\:{B}: \\ $$$$\left({x}−{x}_{{B}} \right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$ \\ $$$${intersection}\:{of}\:{circle}\:{B}\:{and}\:{ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{r}^{\mathrm{2}} −\left({x}−{x}_{{B}} \right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} −{x}_{{B}} ^{\mathrm{2}} +\mathrm{2}{x}_{{B}} {x}}{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{x}^{\mathrm{2}} +{r}^{\mathrm{2}} −{x}^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{4}{r}\:\mathrm{sin}\:\theta\:{x}={b}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right){x}^{\mathrm{2}} −\mathrm{4}{r}\mathrm{sin}\:\theta\:{x}−\left({r}^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta−{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${due}\:{to}\:{tangency} \\ $$$$\Delta=\mathrm{16}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{4}\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left({r}^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta−{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+{r}^{\mathrm{2}} \left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left(\mathrm{1}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\theta−\frac{{b}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4sin}^{\mathrm{2}} \:\theta+\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left[\mathrm{1}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\theta−\left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} \:\theta−\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left(\mathrm{1}+\mathrm{cos}\:\theta\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{cos}\:\theta=\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{cos}\:\theta \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{\mathrm{cos}\:\theta} \\ $$$${let}\:{P}=\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{r}^{\mathrm{4}} }=\frac{{b}^{\mathrm{4}} }{{r}^{\mathrm{4}} \mathrm{cos}\:\theta}=\frac{\left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\theta\right)^{\mathrm{4}} }{\mathrm{cos}\:\theta} \\ $$$${with}\:\lambda=\mathrm{cos}\:\theta \\ $$$$\Rightarrow{P}=\frac{\left(\mathrm{1}+\mathrm{2}\lambda\right)^{\mathrm{4}} }{\lambda} \\ $$$${Area}\:{of}\:{ellipse}\:=\pi{ab} \\ $$$${minimum}\:{area}\:{of}\:{ellipse}\:{means}\:{also} \\ $$$${minimum}\:{value}\:{of}\:{P},\:{therefore} \\ $$$$\frac{{dP}}{{d}\lambda}=\frac{\mathrm{8}\left(\mathrm{1}+\mathrm{2}\lambda\right)^{\mathrm{3}} }{\lambda}−\frac{\left(\mathrm{1}+\mathrm{2}\lambda\right)^{\mathrm{4}} }{\lambda^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{8}−\frac{\mathrm{1}+\mathrm{2}\lambda}{\lambda}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{6}\lambda=\mathrm{1} \\ $$$$\Rightarrow\lambda=\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{35}}}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\sqrt{\mathrm{35}} \\ $$$$\Rightarrow\angle{AOB}=\mathrm{2}\theta=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{35}} \\ $$
Commented by mr W last updated on 18/Aug/19
$${thanks}\:{sir}! \\ $$$${can}\:{you}\:{please}\:{help}\:{checking}\:{my} \\ $$$${solution}.\:{my}\:{result}\:{differs}\:{from} \\ $$$${the}\:{options}\:{given}\:{in}\:{book}. \\ $$
Commented by Tanmay chaudhury last updated on 18/Aug/19
$${sir}\:{you}\:{are}\:{unique}… \\ $$
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
$${thanks}\:{for}\:{comfirming}\:{sir}! \\ $$$${that}\:{means}\:{the}\:{answer}\:{given}\:{in}\:{book} \\ $$$${is}\:{wrong}. \\ $$