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Question Number 41686 by ajfour last updated on 11/Aug/18

Commented by ajfour last updated on 11/Aug/18

$$Findradiusofcircleintermsofa.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 11/Aug/18

$$waitpls...$$

Answered by ajfour last updated on 12/Aug/18

$$Letintersectionofdirectrixand$$$$axisbetheorigin,theneq.of$$$$parabolais$$$$y=\frac{x^2}{4a}+a$$$$eq.ofupperhalfcircle$$$$y=r+\sqrt{r^2-{(x-r)}^2}$$$$solvingthetwo$$$$\frac{x^2}{4a}+a=r+\sqrt{x(2r-x)}$$$$letx=r-r\sin \theta ;then$$$$\frac{r^2{(1-\sin \theta )}^2}{4a}+a=r+r\cos \theta ..\left(i\right)$$$$and\frac{dy}{dx}=\tan \theta =\frac{x}{2a}$$$$\Rightarrow \tan \theta =\frac{r(1-\sin \theta )}{2a}....\left(ii\right)$$$$using\left(ii\right)in\left(i\right):$$$$\tan {}^2\theta +1=\frac{r}{a}(1+\cos \theta )$$$$\Rightarrow (1+\cos \theta )\cos {}^2\theta =\frac{a}{r}...\left(iii\right)$$$$nowusing\left(ii\right)again$$$$(1+\cos \theta )\cos {}^2\theta =\frac{(1-\sin \theta )\cos \theta }{2\sin \theta }$$$$\Rightarrow 2(1+\cos \theta )\sin \theta \cos \theta =1-\sin \theta$$$$\Rightarrow 2(1+\cos \theta )\tan \theta (1+\sin \theta )=1$$$$let\tan \frac{\theta }{2}=tthen$$$$2(1+\frac{1-t^2}{1+t^2})\left(\frac{2t}{1-t^2}\right)(1+\frac{2t}{1+t^2})=1$$$$\Rightarrow 8t{(1+t)}^2=1-t^4$$$$\Rightarrow 8t(1+t)=(1+t^2)(1-t)$$$$\Rightarrow 8t+8t^2=1-t+t^2-t^3$$$$or{t}^3+7t^2+9t-1=0$$$$\Rightarrow \mathit{t}\approx 0.102775(the+veroot)$$$$knowing\mathit{t}weuseeq.\left(iii\right)$$$$r=\frac{a}{(1+\cos \theta )\cos {}^2\theta }$$$$\mathit{r}=\frac{a}{(1+\frac{1-t^2}{1+t^2}){\left(\frac{1-t^2}{1+t^2}\right)}^2}$$$$\mathit{r}=\frac{\mathit{a}{(1+t^2)}^3}{2{(1-t^2)}^2}\approx 0.527\mathit{a}.$$$$$$

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