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Question Number 53323 by ajfour last updated on 20/Jan/19

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

$$y=\frac{x^2}{a}witha=1$$$$A(0,h)$$$$eqn.ofbigcircle:$$$$ay+{(y-h)}^2=R^2$$$$y^2-(2h-a)y+(h^2-R^2)=0$$$$\mathrm{\Delta }={(2h-a)}^2-4(h^2-R^2)=0$$$$-4ah+a^2+4R^2=0$$$$\Rightarrow h=\frac{a^2+4R^2}{4a}$$$$C(r,k)$$$$k=h+\sqrt{{(R+r)}^2-r^2}=h+\sqrt{R(R+2r)}$$$$x_P=r+r\sin \theta$$$$y_P=k-r\cos \theta$$$$y_P^{\prime }=\frac{2x_P}{a}=\tan \theta$$$$\Rightarrow x_P=\frac{a\tan \theta }{2}$$$$\Rightarrow 2r(1+\sin \theta )=a\tan \theta$$$$\Rightarrow r=\frac{a\tan \theta }{2(1+\sin \theta )}...\left(ii\right)$$$${ay}_P=x_P^2$$$$\Rightarrow a(k-r\cos \theta )=\frac{a^2{\tan }^2\theta }{4}$$$$\Rightarrow 4(k-r\cos \theta )=a{\tan }^2\theta$$$$\Rightarrow 4(\frac{a^2+4R^2}{4a}+\sqrt{R(R+2r)}-r\cos \theta )=a{\tan }^2\theta$$$$\Rightarrow 1+4{\left(\frac{R}{a}\right)}^2+4\sqrt{\frac{R}{a}(\frac{R}{a}+\frac{\tan \theta }{1+\sin \theta })}-\frac{2\sin \theta }{1+\sin \theta }={\tan }^2\theta$$$$\Rightarrow 4{\left(\frac{R}{a}\right)}^2+4\sqrt{\frac{R}{a}(\frac{R}{a}+\frac{\tan \theta }{1+\sin \theta })}+\frac{1-\sin \theta }{1+\sin \theta }={\tan }^2\theta ...\left(i\right)$$$$\Rightarrow \theta =\theta \left(\frac{R}{a}\right)....$$$$\Rightarrow r=.....$$$$$$$$example:$$$$R=2\Rightarrow \theta =79.3441°\Rightarrow r=1.3402$$

Commented by ajfour last updated on 20/Jan/19

$$ThanksSir,verylucentmanner!$$

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