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Question Number 89849 by ajfour last updated on 19/Apr/20

Commented by ajfour last updated on 19/Apr/20

$$Findmaximumareaof△PQR$$$$intermsofellipseparametersa,b.$$$$orfindsideof△PQRwhen$$$$itisequilateral.$$

Answered by mr W last updated on 19/Apr/20

$$PARTI$$$$r=b$$$$P{}^{\prime }(r\cos \theta ,r\sin \theta )$$$$Q{}^{\prime }(r\sin \theta ,r\cos \theta )$$$$A=\frac{r^2}{2}\tan (\frac{\pi }{4}+\frac{\theta }{2}){(\cos \theta -\sin \theta )}^2$$$$A=\frac{r^2}{2}\times \frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}{\left(\frac{1-{\tan }^2\frac{\theta }{2}-2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\right)}^2$$$$lett=\tan \frac{\theta }{2}$$$$\Rightarrow \frac{2A}{r^2}=\frac{1+\mathrm{t}}{1-\mathrm{t}}{\left(\frac{1-t^2-2t}{1+t^2}\right)}^2$$$$\frac{d}{dt}\left(\frac{2A}{r^2}\right)=0$$$$.....$$$$\Rightarrow t=-0.1526$$$$\Rightarrow \theta =-17.3528°$$$$A_{max}=\frac{1.1538r^2}{2}=0.5769r^2=0.5769ab$$

Commented by mr W last updated on 19/Apr/20

Commented by ajfour last updated on 20/Apr/20

$$sirwhy{x}_{P^{\prime }}=y_{Q^{\prime }}\mathrm{\&}{y}_{P^{\prime }}=x_{Q^{\prime }}$$$$andwheredoes{}^{\prime }{a}^{\prime }go?$$

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

$$incaseofcircleweusethesymmetry$$$$forthemax.area.$$$$incaseofellipseweonlyneedto$$$$stretchthecircleinx-directionin$$$$ratio\frac{a}{r},i.e.\frac{a}{b}.$$

Answered by mr W last updated on 19/Apr/20

Commented by mr W last updated on 19/Apr/20

$$PARTII$$$$thepointRmustlieonthegreencircle$$$$withcenteratC.$$$$let\alpha ={\tan }^{-1}\frac{b}{a},\mu =\frac{b}{a}$$$$AC=\frac{AB}{2\cos 30°}=\sqrt{\frac{a^2+b^2}{3}}$$$$x_C=-a+AC\cos (30°-\alpha )$$$$x_C=-a+\sqrt{\frac{a^2+b^2}{3}}(\frac{\sqrt{3}}{2}\times \frac{a}{\sqrt{a^2+b^2}}+\frac{1}{2}\times \frac{b}{\sqrt{a^2+b^2}})$$$$\Rightarrow x_C=-\frac{1}{2}(a-\frac{b}{\sqrt{3}})$$$$y_C=AC\sin (30°-\alpha )=\sqrt{\frac{a^2+b^2}{3}}(\frac{1}{2}\times \frac{a}{\sqrt{a^2+b^2}}-\frac{\sqrt{3}}{2}\times \frac{b}{\sqrt{a^2+b^2}})$$$$\Rightarrow y_C=\frac{1}{2}(\frac{a}{\sqrt{3}}-b)$$$$$$$$eqn.ofgreencircle:$$$${[x+\frac{1}{2}(a-\frac{b}{\sqrt{3}})]}^2+{[y-\frac{1}{2}(\frac{a}{\sqrt{3}}-b)]}^2=\frac{a^2+b^2}{3}$$$$or$$$$x_R=-\frac{1}{2}(a-\frac{b}{\sqrt{3}})+\sqrt{\frac{a^2+b^2}{3}}\cos \theta$$$$y_R=\frac{1}{2}(\frac{a}{\sqrt{3}}-b)+\sqrt{\frac{a^2+b^2}{3}}\sin \theta$$$$\Rightarrow \frac{x_R}{a}=-\frac{1}{2}(1-\frac{\mu }{\sqrt{3}})+\sqrt{\frac{1+{\mu }^2}{3}}\cos \theta$$$$\Rightarrow \frac{y_R}{a}=\frac{1}{2}(\frac{1}{\sqrt{3}}-\mu )+\sqrt{\frac{1+{\mu }^2}{3}}\sin \theta$$$$$$$$eqn.ofAR:$$$$y=\frac{a-\sqrt{3}b+2\sqrt{a^2+b^2}\sin \theta }{\sqrt{3}a+b+2\sqrt{a^2+b^2}\cos \theta }(x+a)$$$$y=\frac{1-\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }{\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }(x+a)$$$$intersectionpointP:$$$$y_P=\frac{1-\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }{\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }(x_P+a)$$$${\left(\frac{x_P}{a}\right)}^2+{\left(\frac{y_P}{b}\right)}^2=1$$$${\left(\frac{1-\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }{\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }\right)}^2{(1+\frac{x_P}{a})}^2={\mu }^2[1-{\left(\frac{x_P}{a}\right)}^2]$$$${\left(\frac{1-\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }{\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }\right)}^2(1+\frac{x_P}{a})={\mu }^2(1-\frac{x_P}{a})$$$$\Rightarrow \frac{x_P}{a}=\frac{{\mu }^2-{\left(\frac{1-\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }{\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }\right)}^2}{{\mu }^2+{\left(\frac{1-\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }{\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }\right)}^2}$$$$\Rightarrow \frac{y_P}{a}=\left(\frac{1-\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }{\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }\right)(1+\frac{x_P}{a})$$$$\frac{{PR}^2}{a^2}={(\frac{x_P}{a}-\frac{x_R}{a})}^2+{(\frac{y_P}{a}-\frac{y_R}{a})}^2$$$$$$$$eqn.ofBR:$$$$\frac{y+b}{y_R+b}=\frac{x}{x_R}$$$$y=-b+\frac{a+\sqrt{3}b+2\sqrt{a^2+b^2}\sin \theta }{-\sqrt{3}a+b+2\sqrt{a^2+b^2}\cos \theta }x$$$$y=-b+\frac{1+\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }{-\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }x$$$$intersectionpointQ:$$$$y_Q=-b+\frac{1+\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }{-\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }{x}_Q$$$$x_Q=\frac{-\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }{1+\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }(y_Q+b)$$$${\left(\frac{x_Q}{a}\right)}^2+{\left(\frac{y_Q}{b}\right)}^2=1$$$${\mu }^2{\left(\frac{-\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }{1+\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }\right)}^2{(\frac{y_Q}{a}+\mu )}^2+{\left(\frac{y_Q}{a}\right)}^2={\mu }^2$$$${\mu }^2{\left(\frac{-\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }{1+\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }\right)}^2(\frac{y_Q}{a}+\mu )=\mu -\frac{y_Q}{a}$$$$\Rightarrow \frac{y_Q}{a}=\frac{\mu [1-{\mu }^2{\left(\frac{-\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }{1+\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }\right)}^2]}{1+{\mu }^2{\left(\frac{-\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }{1+\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }\right)}^2}$$$$\Rightarrow \frac{x_Q}{a}=\left(\frac{-\sqrt{3}+\mu +2\sqrt{1+{\mu }^2}\cos \theta }{1+\sqrt{3}\mu +2\sqrt{1+{\mu }^2}\sin \theta }\right)(\frac{y_Q}{a}+\mu )$$$$\frac{{QR}^2}{a^2}={(\frac{x_Q}{a}-\frac{x_R}{a})}^2+{(\frac{y_Q}{a}-\frac{y_R}{a})}^2$$$$$$$$suchthat\mathrm{\Delta }PQRisequilateral,$$$$PR=QR$$$$\frac{{PR}^2}{a^2}=\frac{{QR}^2}{a^2}$$

Commented by mr W last updated on 19/Apr/20

$${(\frac{x_P}{a}-\frac{x_R}{a})}^2+{(\frac{y_P}{a}-\frac{y_R}{a})}^2={(\frac{x_Q}{a}-\frac{x_R}{a})}^2+{(\frac{y_Q}{a}-\frac{y_R}{a})}^2$$$$$$$$fromthiseqn.weget\theta forthe$$$$positionofpointR.$$

Commented by mr W last updated on 19/Apr/20

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

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

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

Commented by mr W last updated on 21/Apr/20

Answered by ajfour last updated on 20/Apr/20

Commented by ajfour last updated on 20/Apr/20

$$letR(h,k)$$$$Eq.AP:y=\frac{k(x+a)}{h+a}$$$$b^2{x}^2+\frac{k^2{(x+a)}^2{a}^2}{{(h+a)}^2}=a^2{b}^2$$$${(h+a)}^2{b}^2{x}^2+k^2{a}^2{(x+a)}^2=a^2{b}^2{(h+a)}^2$$$$rootsare{x}_P,-a$$$$\Rightarrow -{ax}_P=\frac{a^2[a^2{k}^2-b^2{(h+a)}^2]}{b^2{(h+a)}^2+k^2{a}^2}$$$$x_P=\frac{a[b^2{(h+a)}^2-a^2{k}^2]}{b^2{(h+a)}^2+k^2{a}^2}...\left(I\right)$$$$Eq.ofBQ:$$$$y=\left(\frac{k+b}{h}\right)x-b$$$$b^2{h}^2{x}^2+a^2{[(k+b)x-bh]}^2=a^2{b}^2{h}^2$$$$rootsarex=0,x_Q$$$$x_Q=\frac{2a^{2}bh(k+b)}{b^2{h}^2+a^2{(k+b)}^2}...\left(II\right)$$$$\frac{\frac{k+b}{h}-\frac{k}{h+a}}{1+\left(\frac{k+b}{h}\right)\left(\frac{k}{h+a}\right)}=\sqrt{3}....\left(III\right)$$$$slopeofPQ:$$$$\frac{y_Q-y_P}{x_P-x_Q}=\frac{\sqrt{3}-\frac{k}{h+a}}{1+\sqrt{3}\left(\frac{k}{h+a}\right)}...\left(IV\right)$$$$let\frac{k}{h+a}=p,\frac{k+b}{h}=q$$$$x_P=\frac{a[b^2{(h+a)}^2-a^2{k}^2]}{b^2{(h+a)}^2+k^2{a}^2}$$$$=\frac{a(b^2-a^2{p}^2)}{b^2+a^2{p}^2}$$$$x_Q=\frac{2a^{2}bh(k+b)}{b^2{h}^2+a^2{(k+b)}^2}$$$$=\frac{2a^{2}bq}{b^2+a^2{q}^2}$$$$y_P=p(x_P+a)=\frac{p\left(2{ab}^2\right)}{b^2+a^2{p}^2}$$$$y_Q={qx}_Q-b=\frac{2a^2{bq}^2}{b^2+a^2{q}^2}-b$$$$=\frac{b(a^2{q}^2-b^2)}{b^2+a^2{q}^2}$$$$\frac{\frac{b(a^2{q}^2-b^2)}{b^2+a^2{q}^2}-\frac{p\left(2{ab}^2\right)}{b^2+a^2{p}^2}}{\frac{a(b^2-a^2{p}^2)}{b^2+a^2{p}^2}-\frac{2a^{2}bq}{b^2+a^2{q}^2}}=\frac{\sqrt{3}-p}{1+p\sqrt{3}}$$$$......\left(i\right)$$$$And\frac{q-p}{1+pq}=\sqrt{3}.......\left(ii\right)$$$$\frac{b(a^2{q}^2-b^2)(b^2+a^2{p}^2)-2{ab}^{2}p(b^2+a^2{q}^2)}{a(b^2-a^2{p}^2)(b^2+a^2{q}^2)+2a^{2}bq(b^2+a^2{p}^2)}$$$$=\frac{\sqrt{3}-p}{1+p\sqrt{3}}$$$$....$$

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

$$youcangetalsoafinaleqn.withone$$$$unknowninthisway.$$

Commented by ajfour last updated on 21/Apr/20

$$Ihaveabetterwaytoo,Sir,$$$$butletsleaveithere;you$$$$solveditwonderfully,both$$$$parts,\mathcal{T}hanksalot!$$

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