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Question Number 87413 by ajfour last updated on 04/Apr/20

Commented by ajfour last updated on 04/Apr/20

Express r_(min) in terms of a, b .

$${Express}\:{r}_{{min}} \:{in}\:{terms}\:{of}\:{a},\:{b}\:. \\ $$

Answered by mr W last updated on 04/Apr/20

Commented by mr W last updated on 04/Apr/20

M(r cos θ, r sin θ) MC=r ((((r cos θ)/a)+((r sin θ)/b)−1)/(√((1/a^2 )+(1/b^2 ))))=−r ⇒(1/r)=((cos θ)/a)+((sin θ)/b)+(√((1/a^2 )+(1/b^2 ))) ((d((1/r)))/dθ)=−((sin θ)/a)+((cos θ)/b)=0 ⇒tan θ=(a/b) ⇒sin θ=(a/(√(a^2 +b^2 ))), cos θ=(b/(√(a^2 +b^2 ))) ⇒((1/r))_(max) =(b/(a(√(a^2 +b^2 ))))+(a/(b(√(a^2 +b^2 ))))+(√((1/a^2 )+(1/b^2 ))) =((2(√(a^2 +b^2 )))/(ab)) ⇒r_(min) =((ab)/(2(√(a^2 +b^2 ))))

$${M}\left({r}\:\mathrm{cos}\:\theta,\:{r}\:\mathrm{sin}\:\theta\right) \\ $$$${MC}={r} \\ $$$$\frac{\frac{{r}\:\mathrm{cos}\:\theta}{{a}}+\frac{{r}\:\mathrm{sin}\:\theta}{{b}}−\mathrm{1}}{\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }}}=−{r} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{r}}=\frac{\mathrm{cos}\:\theta}{{a}}+\frac{\mathrm{sin}\:\theta}{{b}}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }} \\ $$$$\frac{{d}\left(\frac{\mathrm{1}}{{r}}\right)}{{d}\theta}=−\frac{\mathrm{sin}\:\theta}{{a}}+\frac{\mathrm{cos}\:\theta}{{b}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{{a}}{{b}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{{a}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }},\:\mathrm{cos}\:\theta=\frac{{b}}{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{{r}}\right)_{{max}} =\frac{{b}}{{a}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}+\frac{{a}}{{b}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}+\sqrt{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}{{ab}} \\ $$$$\Rightarrow{r}_{{min}} =\frac{{ab}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }} \\ $$

Commented by ajfour last updated on 04/Apr/20

Amazingly Brilliant solution Sir, and its surprising that tan θ=(a/b) and not (b/a). Thank you Sir!

$$\mathcal{A}{mazingly}\:\mathcal{B}{rilliant}\:{solution} \\ $$$${Sir},\:{and}\:{its}\:{surprising}\:{that} \\ $$$$\mathrm{tan}\:\theta=\frac{{a}}{{b}}\:\:{and}\:{not}\:\frac{{b}}{{a}}. \\ $$$${Thank}\:{you}\:{Sir}! \\ $$

Commented by mr W last updated on 04/Apr/20

it′s indeed surprising with tan θ=(a/b), sir!

$${it}'{s}\:{indeed}\:{surprising}\:{with}\:\mathrm{tan}\:\theta=\frac{{a}}{{b}},\:{sir}! \\ $$

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