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Question Number 54229 by ajfour last updated on 31/Jan/19

Commented by ajfour last updated on 31/Jan/19

$$Findmaximuminradiusofcircle$$$$intermsofellipseparametersa,b.$$

Answered by mr W last updated on 01/Feb/19

$$P(a\cos \theta ,b\sin \theta )$$$$AB=\sqrt{a^2+b^2}$$$$eqn.ofAB:$$$$bx+ay+ab=0$$$$d_{\mathrm{\perp }}fromPtoAB=\frac{ab(\cos \theta +\sin \theta +1)}{\sqrt{a^2+b^2}}$$$$AP=\sqrt{a^2{(\cos \theta +1)}^2+b^2{\sin }^2\theta }$$$$BP=\sqrt{a^2{\cos }^2\theta +b^2{(\sin \theta +1)}^2}$$$$A_{\mathrm{\Delta }}=\frac{r}{2}(\sqrt{a^2+b^2}+\sqrt{a^2{(\cos \theta +1)}^2+b^2{\sin }^2\theta }+\sqrt{a^2{\cos }^2\theta +b^2{(\sin \theta +1)}^2})=\frac{1}{2}\times \sqrt{a^2+b^2}\times \frac{ab(\cos \theta +\sin \theta +1)}{\sqrt{a^2+b^2}}$$$$r(\sqrt{a^2+b^2}+\sqrt{a^2{(\cos \theta +1)}^2+b^2{\sin }^2\theta }+\sqrt{a^2{\cos }^2\theta +b^2{(\sin \theta +1)}^2})=ab(\cos \theta +\sin \theta +1)$$$$\Rightarrow r=\frac{ab(\cos \theta +\sin \theta +1)}{\sqrt{a^2+b^2}+\sqrt{a^2{(\cos \theta +1)}^2+b^2{\sin }^2\theta }+\sqrt{a^2{\cos }^2\theta +b^2{(\sin \theta +1)}^2})}$$$$with\lambda =\frac{b}{a}$$$$\Rightarrow \frac{r}{b}=\frac{\cos \theta +\sin \theta +1}{\sqrt{1+{\lambda }^2}+\sqrt{{(\cos \theta +1)}^2+{\lambda }^2{\sin }^2\theta }+\sqrt{{\cos }^2\theta +{\lambda }^2{(\sin \theta +1)}^2})}$$$$\Rightarrow S=\frac{b}{r}=\frac{\sqrt{1+{\lambda }^2}+\sqrt{{(\cos \theta +1)}^2+{\lambda }^2{\sin }^2\theta }+\sqrt{{\cos }^2\theta +{\lambda }^2{(\sin \theta +1)}^2}}{\cos \theta +\sin \theta +1}$$$$\frac{dS}{d\theta }=0$$$$-\frac{[\sqrt{1+{\lambda }^2}+\sqrt{{(\cos \theta +1)}^2+{\lambda }^2{\sin }^2\theta }+\sqrt{{\cos }^2\theta +{\lambda }^2{(\sin \theta +1)}^2}](\cos \theta -\sin \theta )}{{(\cos \theta +\sin \theta +1)}^2}+\frac{1}{\cos \theta +\sin \theta +1}[\frac{-(\cos \theta +1)\sin \theta +{\lambda }^2\sin \theta \cos \theta }{\sqrt{{(\cos \theta +1)}^2+{\lambda }^2{\sin }^2\theta }}+\frac{-\cos \theta \sin \theta +{\lambda }^2(\sin \theta +1)\cos \theta }{\sqrt{{\cos }^2\theta +{\lambda }^2{(\sin \theta +1)}^2}}]=0$$$$\Rightarrow \frac{[\sqrt{1+{\lambda }^2}+\sqrt{{(\cos \theta +1)}^2+{\lambda }^2{\sin }^2\theta }+\sqrt{{\cos }^2\theta +{\lambda }^2{(\sin \theta +1)}^2}](\cos \theta -\sin \theta )}{\cos \theta +\sin \theta +1}=[\frac{({\lambda }^2-1)\sin \theta \cos \theta -\sin \theta }{\sqrt{{(\cos \theta +1)}^2+{\lambda }^2{\sin }^2\theta }}+\frac{({\lambda }^2-1)\cos \theta \sin \theta +{\lambda }^2\cos \theta }{\sqrt{{\cos }^2\theta +{\lambda }^2{(\sin \theta +1)}^2}}]$$$$\Rightarrow \theta =....$$$$examples:$$$$\lambda =\frac{b}{a}=1\Rightarrow \theta =45°$$$$\lambda =\frac{b}{a}=0.5\Rightarrow \theta =71.35°$$$$\lambda =\frac{b}{a}=2\Rightarrow \theta =18.65°$$

Commented by ajfour last updated on 01/Feb/19

$$yesSir,noshorteralternative!$$$$onlyitook\tan \frac{\theta }{2}=m,igot$$$$S=\frac{2b}{r}=\frac{\sqrt{1+{\lambda }^2{m}^2}}{1+m}+\frac{(1-m)\sqrt{{\lambda }^2+m^2}}{m}+\frac{(1+m^2)\sqrt{1+{\lambda }^2}}{1+m}$$$$m\left(\lambda \right)isquiteimplicitupon\frac{dS}{dm}=0.$$$$ThankyouSir.$$

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