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

Commented by ajfour last updated on 04/Apr/20

$$Express{r}_{min}intermsofa,b.$$

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

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

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

Commented by ajfour last updated on 04/Apr/20

$$\mathcal{A}mazingly\mathcal{B}rilliantsolution$$$$Sir,anditssurprisingthat$$$$\tan \theta =\frac{a}{b}andnot\frac{b}{a}.$$$$ThankyouSir!$$

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

$${it}^{\prime }sindeedsurprisingwith\tan \theta =\frac{a}{b},sir!$$

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