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Question Number 50898 by ajfour last updated on 21/Dec/18

Commented by ajfour last updated on 21/Dec/18

$$IflengthofBPismaximum$$$$andequalto\mathit{l},andthetwocoloured$$$$areasequal,find\mathit{a}and\mathit{b}ofellipse.$$

Answered by ajfour last updated on 21/Dec/18

$$igota=l\sqrt{2-\sqrt{2}},b=l\sqrt{3\sqrt{2}-4}.$$

Commented by mr W last updated on 22/Dec/18

$$igotthesamesir.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Dec/18

$$areasky=\frac{1}{2}\times acos\theta \times bsin\theta +{\int }_{bsin\theta }^b\frac{a}{b}\sqrt{b^2-y^2}dy$$$$=\frac{a}{b}[\mid \frac{y}{2}\sqrt{b^2-y^2}+\frac{b^2}{2}{sin}^{-1}\left(\frac{y}{b}\right){\mid }_{bsin\theta }^b]+\frac{1}{4}sin2\theta$$$$=\frac{a}{b}[\left(\frac{b^2}{2}\times \frac{\pi }{2}\right)-(\frac{bsin\theta \times bcos\theta }{2}+\frac{b^2}{2}\theta )]+\frac{sin2\theta }{4}$$$$=\frac{\pi ab}{4}-\frac{absin2\theta }{4}+\frac{ab\theta }{2}+\frac{sin2\theta }{4}$$$$given{A}_s=A_p=\frac{\pi ab}{8}at\theta ={\theta }_0$$$$\frac{\pi ab}{8}=\frac{\pi ab}{4}-\frac{absin2{\theta }_0}{4}+\frac{ab{\theta }_0}{2}+\frac{sin2{\theta }_0}{4}$$$$sin2{\theta }_0\times \frac{1}{4}(ab-1)-\frac{ab}{2}{sin}^{-1}\left(\frac{b^2}{a^2-b^2}\right)=\frac{\pi ab}{8}$$$$$$$$$$$$$$$$$$$$lengthBP=\sqrt{{(acos\theta -0)}^2+{(bsin\theta +b)}^2}=s$$$$s_{max}=lgiven$$$$s^2=a^2{cos}^2\theta +b^2{(1+sin\theta )}^2$$$$2s\times \frac{ds}{d\theta }=a^2\times -sin2\theta +b^2\times 2(1+sin\theta )\left(cos\theta \right)$$$$2b^{2}cos\theta +b^{2}sin2\theta -a^{2}sin2\theta =0$$$$cos\theta [2b^2+2b^{2}sin\theta -2a^{2}sin\theta ]=0$$$$cos\theta =0\theta =\frac{\pi }{2}\leftarrow ignored$$$$sin{\theta }_0=\frac{b^2}{a^2-b^2}cos{\theta }_0=\frac{\sqrt{{(a^2-b^2)}^2-b^4}}{(a^2-b^2)}=\frac{\sqrt{a^4-2a^2{b}^2}}{(a^2-b^2)}$$$$wait...$$$$$$$$$$$$$$$$$$

Answered by mr W last updated on 22/Dec/18

Commented by OTCHRRE ABDULLAI last updated on 22/Dec/18

$$Mymanyouaretoogood$$$$pleaseyouarefromwhichcountry?$$

Commented by mr W last updated on 22/Dec/18

$$r^2=\frac{a^2{b}^2}{a^2{\sin }^2\theta +b^2{\cos }^2\theta }$$$$A_{red}=A_{blue}=\frac{\pi ab}{8}$$$$A_{red}={\int }_0^{\theta }\frac{r^{2}d\theta }{2}=\frac{a^2{b}^2}{2}{\int }_0^{\theta }\frac{d\theta }{a^2{\sin }^2\theta +b^2{\cos }^2\theta }$$$$=\frac{a^2{b}^2}{2}\times \frac{{\tan }^{-1}\left(\frac{a}{b}\tan \theta \right)}{ab}$$$$=\frac{ab}{2}\times {\tan }^{-1}\left(\frac{a}{b}\tan \theta \right)=\frac{\pi ab}{8}$$$$\Rightarrow {\tan }^{-1}\left(\frac{a}{b}\tan \theta \right)=\frac{\pi }{4}$$$$\Rightarrow \frac{a}{b}\tan \theta =1$$$$\Rightarrow \tan \theta =\frac{b}{a}\Rightarrow \sin \theta =\frac{b}{\sqrt{a^2+b^2}}and\cos \theta =\frac{a}{\sqrt{a^2+b^2}}$$$$P(r,\theta )$$$$r^2=\frac{a^2{b}^2}{a^2{\sin }^2\theta +b^2{\cos }^2\theta }=\frac{a^2{b}^2}{\frac{2a^2{b}^2}{a^2+b^2}}=\frac{a^2+b^2}{2}$$$$\frac{x}{a^2}+\frac{y}{b^2}{y}^{\prime }=0$$$$\frac{r\cos \theta }{a^2}+\frac{r\sin \theta }{b^2}{y}^{\prime }=0$$$$\Rightarrow y^{\prime }\left(atP\right)=-\frac{b^2}{a^2\tan \theta }=-\frac{b}{a}=-\tan \alpha =-\tan \theta$$$$\Rightarrow \alpha =\theta$$$$\phi =\frac{\pi }{2}-\alpha =\frac{\pi }{2}-\theta$$$$r\cos \theta =l\cos \phi =l\sin \theta$$$$\Rightarrow r=l\tan \theta =\frac{lb}{a}$$$$r\sin \theta =l\sin \phi -b=l\cos \theta -b$$$$\Rightarrow r=\frac{l}{\tan \theta }-\frac{b}{\sin \theta }$$$$\Rightarrow \frac{l}{\tan \theta }-\frac{b}{\sin \theta }=l\tan \theta$$$$\Rightarrow b=l(\frac{1}{\tan \theta }-\tan \theta )\sin \theta =l\times \frac{a^2-b^2}{ab}\times \frac{b}{\sqrt{a^2+b^2}}$$$$\Rightarrow l=\frac{ab\sqrt{a^2+b^2}}{a^2-b^2}$$$$r^2=\frac{l^2{b}^2}{a^2}=\frac{b^2}{a^2}\times \frac{a^2{b}^2(a^2+b^2)}{{(a^2-b^2)}^2}=\frac{b^4(a^2+b^2)}{{(a^2-b^2)}^2}=\frac{a^2+b^2}{2}$$$$\Rightarrow \frac{b^4}{{(a^2-b^2)}^2}=\frac{1}{2}$$$${\left(\sqrt{2}{b}^2\right)}^2-{(a^2-b^2)}^2=0$$$$[(\sqrt{2}+1)b^2-a^2][(\sqrt{2}-1)b^2+a^2]=0$$$$\Rightarrow a^2=(\sqrt{2}+1)b^2$$$$\Rightarrow l=\frac{ab\sqrt{a^2+b^2}}{a^2-b^2}=\frac{ab\sqrt{(2+\sqrt{2})b^2}}{\sqrt{2}{b}^2}=a\sqrt{\frac{2+\sqrt{2}}{2}}$$$$\Rightarrow a=\sqrt{\frac{2}{2+\sqrt{2}}}l=\sqrt{2-\sqrt{2}}l$$$$\Rightarrow b=\frac{a}{\sqrt{\sqrt{2}+1}}=\sqrt{\frac{2-\sqrt{2}}{\sqrt{2}+1}}l=\sqrt{3\sqrt{2}-4}l$$

Commented by ajfour last updated on 22/Dec/18

$$\mathcal{W}\mathcal{O}\mathcal{W}!$$

Commented by mr W last updated on 22/Dec/18

Commented by mr W last updated on 22/Dec/18

$$diagramshowsthecasel=5.$$

Commented by ajfour last updated on 22/Dec/18

$$TooGoodSir,Thanksforconfirming.$$

Commented by OTCHRRE ABDULLAI last updated on 23/Dec/18

$$Youareworldbestmyman$$$$ifmathematicsisfootballyouwill$$$$bemessiorRonaldoireallylike$$$$youinfactyouarenowmybest$$$$friend$$

Commented by peter frank last updated on 22/Dec/18

$$andphysicstoo$$

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