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Question Number 175836 by ajfour last updated on 08/Sep/22

Commented by ajfour last updated on 08/Sep/22

$$Ellipse:x^2+\frac{y^2}{b^2}=1$$

Answered by mr W last updated on 08/Sep/22

Commented by mr W last updated on 08/Sep/22

$$a=1$$$$\mu =\frac{b}{a}$$$$s=2a$$$$r=\frac{a}{2}$$$$sayP(a\cos \theta ,b\sin \theta )$$$$\tan \varphi =-\frac{dx}{dy}=-\frac{-a\sin \theta }{b\cos \theta }=\frac{\tan \theta }{\mu }$$$$a\cos \theta +r\cos \varphi =r$$$$a\cos \theta =\frac{a}{2}(1-\frac{\mu }{\sqrt{{\mu }^2+{\tan }^2\theta }})$$$$\frac{\mu }{\sqrt{{\mu }^2+{\tan }^2\theta }}=1-2\cos \theta$$$$\Rightarrow \mu =\frac{\tan \theta -2\sin \theta }{2\sqrt{\cos \theta (1-\cos \theta )}}...\left(i\right)$$$$s=b+b\sin \theta +r\sin \varphi +r=2a$$$$\Rightarrow 2\mu (1+\sin \theta )+\frac{\tan \theta }{\sqrt{{\mu }^2+{\tan }^2\theta }}=3...\left(ii\right)$$$$\Rightarrow \theta \approx 1.1693$$$$\Rightarrow \mu \approx 0.5272✓$$

Commented by Tawa11 last updated on 08/Sep/22

$$\mathrm{Great}\mathrm{sir}$$

Commented by mr W last updated on 09/Sep/22

Answered by ajfour last updated on 09/Sep/22

Commented by ajfour last updated on 10/Sep/22

$${(1-\cos \theta )}^2+\frac{{(3-p-\sin \theta )}^2}{p^2}=1..\left(i\right)$$$$And\frac{dx}{dy}=\frac{4y}{p^{2}x}$$$$\Rightarrow \tan \theta =\frac{4(3-p-\sin \theta )}{p^2(1-\cos \theta )}...\left(ii\right)$$$$\Rightarrow {(1-\cos \theta )}^2(1+p^2\tan {}^2\theta )=1$$$$.\Rightarrow \frac{p^2}{4}\tan {}^2\theta =\frac{1}{{(1-\cos \theta )}^2}-1$$$$.....$$

Commented by mr W last updated on 09/Sep/22

$$shoulditnotbe:$$$$\frac{{(1-\cos \theta )}^2}{2^2}+\frac{{(3-p-\sin \theta )}^2}{p^2}=1..\left(i\right)$$$$And\frac{dx}{dy}=\frac{2^{2}y}{p^{2}x}$$

Commented by ajfour last updated on 09/Sep/22

$$yez,thankssir.$$

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