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Question Number 88613 by ajfour last updated on 11/Apr/20

Commented by ajfour last updated on 11/Apr/20

Find a/b .

$${Find}\:{a}/{b}\:. \\ $$

Commented by ajfour last updated on 11/Apr/20

(circle should have been bit larger)

$$\left({circle}\:{should}\:{have}\:{been}\:{bit}\:{larger}\right) \\ $$

Commented by ajfour last updated on 11/Apr/20

sir i got (a/b)= (((√5)+1)/2) .

$${sir}\:{i}\:{got}\:\frac{{a}}{{b}}=\:\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\:. \\ $$

Answered by mr W last updated on 11/Apr/20

intersection of circle and ellipse at point P(a cos θ, b sin θ) ((a cos θ)/(b+b sin θ))=((b cos θ)/(a sin θ)) with μ=(b/a) (1/((1+sin θ)))=(μ^2 /(sin θ)) ⇒sin θ=(μ^2 /(1−μ^2 )) (μ≠1) (a−b sin θ)(b+b sin θ)=a^2 cos^2 θ μ(1−μ sin θ)=1−sin θ μ(1−(μ^3 /(1−μ^2 )))=1−(μ^2 /(1−μ^2 )) ⇒μ^4 +μ^3 −2μ^2 −μ+1=0 ⇒(μ−1)(μ+1)(μ^2 +μ−1)=0 ⇒μ=(((√5)−1)/2)

$${intersection}\:{of}\:{circle}\:{and}\:{ellipse}\:{at} \\ $$$${point}\:{P}\left({a}\:\mathrm{cos}\:\theta,\:{b}\:\mathrm{sin}\:\theta\right) \\ $$$$\frac{{a}\:\mathrm{cos}\:\theta}{{b}+{b}\:\mathrm{sin}\:\theta}=\frac{{b}\:\mathrm{cos}\:\theta}{{a}\:\mathrm{sin}\:\theta} \\ $$$${with}\:\mu=\frac{{b}}{{a}} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{sin}\:\theta\right)}=\frac{\mu^{\mathrm{2}} }{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\mu^{\mathrm{2}} }{\mathrm{1}−\mu^{\mathrm{2}} }\:\:\:\:\:\left(\mu\neq\mathrm{1}\right) \\ $$$$\left({a}−{b}\:\mathrm{sin}\:\theta\right)\left({b}+{b}\:\mathrm{sin}\:\theta\right)={a}^{\mathrm{2}} \:\mathrm{cos}^{\mathrm{2}} \:\theta \\ $$$$\mu\left(\mathrm{1}−\mu\:\mathrm{sin}\:\theta\right)=\mathrm{1}−\mathrm{sin}\:\theta \\ $$$$\mu\left(\mathrm{1}−\frac{\mu^{\mathrm{3}} }{\mathrm{1}−\mu^{\mathrm{2}} }\right)=\mathrm{1}−\frac{\mu^{\mathrm{2}} }{\mathrm{1}−\mu^{\mathrm{2}} } \\ $$$$\Rightarrow\mu^{\mathrm{4}} +\mu^{\mathrm{3}} −\mathrm{2}\mu^{\mathrm{2}} −\mu+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mu−\mathrm{1}\right)\left(\mu+\mathrm{1}\right)\left(\mu^{\mathrm{2}} +\mu−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mu=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}} \\ $$

Commented by ajfour last updated on 11/Apr/20

till sin θ= (b^2 /(a^2 −b^2 )) (i agree sir). then i wrote eq. of tangent to ellipse (x/a)cos θ+(y/b)sin θ=1 its y_(intercept) = radius = a ⇒ (a/b)sin θ=1 ⇒ (a/b)((b^2 /(a^2 −b^2 )))=1 let (a/b)=m ⇒ m=m^2 −1 ⇒ m^2 −m−1=0 m=(1/2)+((√5)/2) .

$${till}\:\mathrm{sin}\:\theta=\:\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:\:\:\left({i}\:{agree}\:{sir}\right). \\ $$$${then}\:{i}\:{wrote}\:{eq}.\:{of}\:{tangent}\:{to} \\ $$$${ellipse} \\ $$$$\:\:\frac{{x}}{{a}}\mathrm{cos}\:\theta+\frac{{y}}{{b}}\mathrm{sin}\:\theta=\mathrm{1} \\ $$$${its}\:{y}_{{intercept}} \:=\:{radius}\:=\:{a} \\ $$$$\Rightarrow\:\:\:\frac{{a}}{{b}}\mathrm{sin}\:\theta=\mathrm{1} \\ $$$$\Rightarrow\:\:\frac{{a}}{{b}}\left(\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$${let}\:\:\frac{{a}}{{b}}={m} \\ $$$$\Rightarrow\:\:\:{m}={m}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\:\:{m}^{\mathrm{2}} −{m}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:{m}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:. \\ $$$$ \\ $$

Commented by mr W last updated on 11/Apr/20

Commented by mr W last updated on 11/Apr/20

it′s a nice question sir!

$${it}'{s}\:{a}\:{nice}\:{question}\:{sir}! \\ $$

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