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 θ=((∠AOB)/2) eqn. of ellipse: (x^2 /a^2 )+(y^2 /b^2 )=1 with b=r+2r cos θ=r(1+2 cos θ) x_B =2r sin θ eqn. of circle at B: (x−x_B )^2 +y^2 =r^2 intersection of circle B and ellipse: (x^2 /a^2 )+((r^2 −(x−x_B )^2 )/b^2 )=1 (x^2 /a^2 )+((r^2 −x^2 −x_B ^2 +2x_B x)/b^2 )=1 (b^2 /a^2 )x^2 +r^2 −x^2 −4r^2 sin^2 θ+4r sin θ x=b^2 (1−(b^2 /a^2 ))x^2 −4rsin θ x−(r^2 −4r^2 sin^2 θ−b^2 )=0 due to tangency Δ=16r^2 sin^2 θ+4(1−(b^2 /a^2 ))(r^2 −4r^2 sin^2 θ−b^2 )=0 ⇒4r^2 sin^2 θ+r^2 (1−(b^2 /a^2 ))(1−4 sin^2 θ−(b^2 /r^2 ))=0 ⇒4sin^2 θ+(1−(b^2 /a^2 ))[1−4 sin^2 θ−(1+2 cos θ)^2 ]=0 ⇒sin^2 θ−(1−(b^2 /a^2 ))(1+cos θ)=0 ⇒1−cos θ=1−(b^2 /a^2 ) ⇒(b^2 /a^2 )=cos θ ⇒a^2 =(b^2 /(cos θ)) let P=((a^2 b^2 )/r^4 )=(b^4 /(r^4 cos θ))=(((1+2 cos θ)^4 )/(cos θ)) with λ=cos θ ⇒P=(((1+2λ)^4 )/λ) Area of ellipse =πab minimum area of ellipse means also minimum value of P, therefore (dP/dλ)=((8(1+2λ)^3 )/λ)−(((1+2λ)^4 )/λ^2 )=0 8−((1+2λ)/λ)=0 ⇒6λ=1 ⇒λ=cos θ=(1/6) ⇒sin θ=((√(35))/6) ⇒tan θ=(√(35)) ⇒∠AOB=2θ=2 tan^(−1) (√(35))](Q66662.png)
$${let}\:{r}={radius}\:{of}\:{the}\:{circles} \\ $$$${let}\:\theta=\frac{\angle{AOB}}{\mathrm{2}} \\ $$$${eqn}.\:{of}\:{ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${with}\:{b}={r}+\mathrm{2}{r}\:\mathrm{cos}\:\theta={r}\left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\theta\right) \\ $$$${x}_{{B}} =\mathrm{2}{r}\:\mathrm{sin}\:\theta \\ $$$${eqn}.\:{of}\:{circle}\:{at}\:{B}: \\ $$$$\left({x}−{x}_{{B}} \right)^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$ \\ $$$${intersection}\:{of}\:{circle}\:{B}\:{and}\:{ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{r}^{\mathrm{2}} −\left({x}−{x}_{{B}} \right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{r}^{\mathrm{2}} −{x}^{\mathrm{2}} −{x}_{{B}} ^{\mathrm{2}} +\mathrm{2}{x}_{{B}} {x}}{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }{x}^{\mathrm{2}} +{r}^{\mathrm{2}} −{x}^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{4}{r}\:\mathrm{sin}\:\theta\:{x}={b}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right){x}^{\mathrm{2}} −\mathrm{4}{r}\mathrm{sin}\:\theta\:{x}−\left({r}^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta−{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${due}\:{to}\:{tangency} \\ $$$$\Delta=\mathrm{16}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{4}\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left({r}^{\mathrm{2}} −\mathrm{4}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta−{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \:\theta+{r}^{\mathrm{2}} \left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left(\mathrm{1}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\theta−\frac{{b}^{\mathrm{2}} }{{r}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4sin}^{\mathrm{2}} \:\theta+\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left[\mathrm{1}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\theta−\left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} \right]=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}^{\mathrm{2}} \:\theta−\left(\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)\left(\mathrm{1}+\mathrm{cos}\:\theta\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{cos}\:\theta=\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{cos}\:\theta \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{\mathrm{cos}\:\theta} \\ $$$${let}\:{P}=\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{r}^{\mathrm{4}} }=\frac{{b}^{\mathrm{4}} }{{r}^{\mathrm{4}} \mathrm{cos}\:\theta}=\frac{\left(\mathrm{1}+\mathrm{2}\:\mathrm{cos}\:\theta\right)^{\mathrm{4}} }{\mathrm{cos}\:\theta} \\ $$$${with}\:\lambda=\mathrm{cos}\:\theta \\ $$$$\Rightarrow{P}=\frac{\left(\mathrm{1}+\mathrm{2}\lambda\right)^{\mathrm{4}} }{\lambda} \\ $$$${Area}\:{of}\:{ellipse}\:=\pi{ab} \\ $$$${minimum}\:{area}\:{of}\:{ellipse}\:{means}\:{also} \\ $$$${minimum}\:{value}\:{of}\:{P},\:{therefore} \\ $$$$\frac{{dP}}{{d}\lambda}=\frac{\mathrm{8}\left(\mathrm{1}+\mathrm{2}\lambda\right)^{\mathrm{3}} }{\lambda}−\frac{\left(\mathrm{1}+\mathrm{2}\lambda\right)^{\mathrm{4}} }{\lambda^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{8}−\frac{\mathrm{1}+\mathrm{2}\lambda}{\lambda}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{6}\lambda=\mathrm{1} \\ $$$$\Rightarrow\lambda=\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{35}}}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\sqrt{\mathrm{35}} \\ $$$$\Rightarrow\angle{AOB}=\mathrm{2}\theta=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{35}} \\ $$
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
$${thanks}\:{for}\:{comfirming}\:{sir}! \\ $$$${that}\:{means}\:{the}\:{answer}\:{given}\:{in}\:{book} \\ $$$${is}\:{wrong}. \\ $$