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Question-66656




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

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