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Question Number 52627 by ajfour last updated on 10/Jan/19

Commented by ajfour last updated on 10/Jan/19

$$Findradiusofsuchacircle.$$

Answered by mr W last updated on 10/Jan/19

$$Q(-h,{ah}^2)$$$$y_Q^{\prime }=-2ah$$$$eqn.ofQP:$$$$y={ah}^2+\frac{(x+h)}{2ah}$$$$y={ax}^2$$$$\Rightarrow {ah}^2+\frac{x+h}{2ah}={ax}^2$$$$\Rightarrow x^2-\frac{x}{2a^{2}h}-h^2-\frac{1}{2a^2}=0$$$$\Rightarrow x_P=\frac{1}{2}[\frac{1}{2a^{2}h}+\sqrt{\frac{1}{4a^4{h}^2}+4(h^2+\frac{1}{2a^2})}]$$$$\Rightarrow x_P=h+\frac{1}{2a^{2}h}$$$$\Rightarrow y_P={ah}^2+\frac{1}{2ah}(2h+\frac{1}{2a^{2}h})$$$$\Rightarrow y_P={ah}^2+\frac{1}{a}+\frac{1}{4a^3{h}^2}$$$${PQ}^2={(h+\frac{1}{2a^{2}h}+h)}^2+{({ah}^2+\frac{1}{a}+\frac{1}{4a^3{h}^2}-{ah}^2)}^2$$$${PQ}^2={(2h+\frac{1}{2a^{2}h})}^2+{(\frac{1}{a}+\frac{1}{4a^3{h}^2})}^2=y_P^2$$$${(2h+\frac{1}{2a^{2}h})}^2+{(\frac{1}{a}+\frac{1}{4a^3{h}^2})}^2={({ah}^2+\frac{1}{a}+\frac{1}{4a^3{h}^2})}^2$$$$\Rightarrow {(2ah+\frac{1}{2ah})}^2=h^2{a}^2(2+a^2{h}^2+\frac{1}{2a^2{h}^2})$$$$let\lambda =ah$$$$\Rightarrow {(2\lambda +\frac{1}{2\lambda })}^2={\lambda }^2(2+{\lambda }^2+\frac{1}{2{\lambda }^2})$$$$\Rightarrow 4{\lambda }^6-8{\lambda }^4-6{\lambda }^2-1=0$$$$\Rightarrow \lambda \approx 1.6159$$$$\Rightarrow R=y_P=\frac{1}{a}(1+{\lambda }^2+\frac{1}{4{\lambda }^2})\approx \frac{3.7069}{a}$$

Commented by mr W last updated on 11/Jan/19

Commented by ajfour last updated on 11/Jan/19

$$VerybrilliantSir!Thankyou.$$

Answered by ajfour last updated on 11/Jan/19

$$Let\mathrm{\angle }QPF=\theta ,Q(-h,{ah}^2)$$$$\frac{dy}{dx}{\mid }_Q=-\tan \theta =-2ah$$$$y_P=R={ah}^2+R\cos \theta$$$$x_P=-h+R\sin \theta$$$$y_P={ax}_P^2$$$$\Rightarrow R={ah}^2+R\cos \theta =a{(R\sin \theta -h)}^2$$$$\Rightarrow \cos \theta =aR\sin {}^2\theta -2ah\sin \theta$$$$\Rightarrow \cos \theta =\frac{a^2{h}^2\sin {}^2\theta }{1-\cos \theta }-2ah\sin \theta$$$$letusnowuseah=\frac{\tan \theta }{2}$$$$\Rightarrow \cos \theta =\frac{\tan {}^2\theta \sin {}^2\theta }{4(1-\cos \theta )}-\sin \theta \tan \theta$$$$\Rightarrow 4(1-\cos \theta )\cos {}^2\theta (\cos \theta +\frac{\sin {}^2\theta }{\cos \theta })=\sin {}^4\theta$$$$let\cos \theta =\lambda$$$$\Rightarrow 4(1-\lambda )\lambda ={(1-{\lambda }^2)}^2$$$$\Rightarrow 4\lambda ={(1+\lambda )}^2(1-\lambda )$$$$\Rightarrow \lambda \approx 0.2956$$$$R=\frac{{ah}^2}{1-\cos \theta }=\frac{\tan {}^2\theta }{4a(1-\cos \theta )}$$$$=\frac{(\frac{1}{\cos {}^2\theta }-1)}{4a(1-\cos \theta )}=\frac{(1+\cos \theta )}{4a\cos {}^2\theta }$$$$R\approx \frac{3.707}{a}.$$

Commented by mr W last updated on 11/Jan/19

$$thankyousir!$$$$canyouget\lambda exactly?{it}^{\prime }sacubiceqn.$$

Commented by ajfour last updated on 12/Jan/19

$$Thankssir,imeantitriedbut$$$${couldn}^{\prime }tmatchwiththecalculated$$$$answerinoneattempt.$$

Commented by ajfour last updated on 11/Jan/19

$$nosir,howcome?$$

Commented by mr W last updated on 11/Jan/19

$$4\lambda ={(1+\lambda )}^2(1-\lambda )$$$$\Rightarrow {\lambda }^3+{\lambda }^2+3\lambda -1=0(cubiceqn.)$$$$let\lambda =t+\delta$$$$t^3+(3\delta +1)t^2+(3{\delta }^2+2\delta +3)t+{\delta }^3+{\delta }^2+3\delta -1=0$$$$let3\delta +1=0\Rightarrow \delta =-\frac{1}{3}$$$$\Rightarrow t^3+\frac{8}{3}t-\frac{52}{27}=0$$$${(u+v)}^3-3uv(u+v)-(u^3+v^3)=0$$$$lett=u+v$$$$-3uv=\frac{8}{3}$$$$\Rightarrow u^3{v}^3=-\frac{512}{729}$$$$\Rightarrow u^3+v^3=\frac{52}{27}$$$$u^{3}and{v}^{3}arerootsof$$$$x^2-\frac{52}{27}x-\frac{512}{729}=0$$$$x=\frac{2(13\pm 3\sqrt{33})}{27}=u^{3}and{v}^3$$$$\Rightarrow u=\frac{\sqrt[3]{2(13+3\sqrt{33})}}{3}$$$$\Rightarrow v=\frac{\sqrt[3]{2(13-3\sqrt{33})}}{3}$$$$\Rightarrow \lambda =t+\delta =u+v+\delta$$$$\Rightarrow \lambda =\frac{\sqrt[3]{2(13+3\sqrt{33})}+\sqrt[3]{2(13-3\sqrt{33})}-1}{3}\approx 0.295598$$

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