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Question Number 89010 by ajfour last updated on 14/Apr/20

Commented by ajfour last updated on 14/Apr/20

$$FindC(h,k)intermsofa,b.$$

Commented by ajfour last updated on 14/Apr/20

$$yessir,Iagree.$$

Commented by mr W last updated on 14/Apr/20

$$ODisnotindependent,soitshould$$$$notbegiven.$$

Commented by ajfour last updated on 15/Apr/20

Answered by mr W last updated on 14/Apr/20

Commented by mr W last updated on 16/Apr/20

$$let\mu =\frac{b}{a}$$$$\frac{r^2{\cos }^2\theta }{a^2}+\frac{r^2{\sin }^2\theta }{b^2}=1$$$$r=\frac{ab}{\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }}$$$$\frac{x}{a^2}+\frac{y}{b^2}\times \frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac{b^{2}x}{a^{2}y}=-\frac{b^2}{a^2\tan \theta }$$$$$$$$CA=CB=\frac{ab}{\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }}$$$$CD=\frac{ab}{\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }}$$$$\tan (\phi -\theta )=\frac{dy}{dx}=\frac{b^2}{a^2\tan \theta }=\frac{{\mu }^2}{\tan \theta }$$$$\frac{\tan \phi -\tan \theta }{1+\tan \phi \tan \theta }=\frac{{\mu }^2}{\tan \theta }$$$$\Rightarrow \tan \phi =\frac{{\mu }^2+{\tan }^2\theta }{(1-{\mu }^2)\tan \theta }$$$$$$$$\tan \phi =\frac{CB}{CD}=\sqrt{\frac{a^2{\cos }^2\theta +b^2{\sin }^2\theta }{a^2{\sin }^2\theta +b^2{\cos }^2\theta }}$$$$\Rightarrow \tan \phi =\sqrt{\frac{1+{\mu }^2{\tan }^2\theta }{{\mu }^2+{\tan }^2\theta }}$$$$$$$$\frac{{\mu }^2+{\tan }^2\theta }{(1-{\mu }^2)\tan \theta }=\sqrt{\frac{1+{\mu }^2{\tan }^2\theta }{{\mu }^2+{\tan }^2\theta }}$$$$\frac{{\mu }^4+2{\mu }^2{\tan }^2\theta +{\tan }^4\theta }{{(1-{\mu }^2)}^2{\tan }^2\theta }=\frac{1+{\mu }^2{\tan }^2\theta }{{\mu }^2+{\tan }^2\theta }$$$${\mu }^6+2{\mu }^4{\tan }^2\theta +{\mu }^2{\tan }^4\theta +{\mu }^4{\tan }^2\theta +2{\mu }^2{\tan }^4\theta +{\tan }^6\theta ={(1-{\mu }^2)}^2{\tan }^2\theta +{\mu }^2{(1-{\mu }^2)}^2{\tan }^4\theta$$$${\tan }^6\theta +{\mu }^2(2+2{\mu }^2-{\mu }^4){\tan }^4\theta -(1-2{\mu }^2-2{\mu }^4){\tan }^2\theta +{\mu }^6=0$$$$({\tan }^2\theta +1)[{\tan }^4\theta -(1+{\mu }^2)(1-3{\mu }^2+{\mu }^4){\tan }^2\theta +{\mu }^6]=0$$$$\Rightarrow {\tan }^2\theta =\frac{(1+{\mu }^2)(1-3{\mu }^2+{\mu }^4)+\sqrt{{(1+{\mu }^2)}^2{(1-3{\mu }^2+{\mu }^4)}^2-4{\mu }^6}}{2}$$$$\Rightarrow \tan \theta =\sqrt{\frac{1-2{\mu }^2-2{\mu }^4+{\mu }^6+(1-{\mu }^2)\sqrt{1-2{\mu }^2-5{\mu }^4-2{\mu }^6+{\mu }^8}}{2}}=\delta$$$$\mathrm{\Delta }=1-2{\mu }^2-5{\mu }^4-2{\mu }^6+{\mu }^8⩾0$$$$\Rightarrow \mu ⩽\sqrt{\frac{1+2\sqrt{2}-\sqrt{5+4\sqrt{2}}}{2}}\approx 0.531$$$$$$$$\tan \phi =\frac{{\mu }^2+{\tan }^2\theta }{(1-{\mu }^2)\tan \theta }=\frac{{\mu }^2+{\delta }^2}{(1-{\mu }^2)\delta }$$$$\Rightarrow \sin \phi =\frac{{\mu }^2+{\delta }^2}{\sqrt{(1+{\delta }^2)({\mu }^4+{\delta }^2)}}$$$$$$$$x_C=h=CD\sin \phi =\frac{ab\sin \phi }{\sqrt{a^2{\cos }^2\theta +b^2{\sin }^2\theta }}$$$$\frac{h}{a}=\frac{\mu \sin \phi }{\sqrt{{\mu }^2+(1-{\mu }^2){\cos }^2\theta }}$$$$\Rightarrow \frac{h}{a}=\frac{\mu ({\mu }^2+{\delta }^2)}{\sqrt{(1+{\mu }^2{\delta }^2)({\mu }^4+{\delta }^2)}}$$$$$$$$y_C=k=CA\sin \phi =\frac{ab\sin \phi }{\sqrt{a^2{\sin }^2\theta +b^2{\cos }^2\theta }}$$$$\frac{k}{a}=\frac{\mu \sin \phi }{\sqrt{1-(1-{\mu }^2){\cos }^2\theta }}$$$$\Rightarrow \frac{k}{a}=\mu \sqrt{\frac{{\mu }^2+{\delta }^2}{{\mu }^4+{\delta }^2}}$$

Commented by mr W last updated on 15/Apr/20

Commented by ajfour last updated on 15/Apr/20

$$Thankyou,Sir.$$

Commented by mr W last updated on 15/Apr/20

Commented by mr W last updated on 15/Apr/20

Commented by ajfour last updated on 16/Apr/20

$$Looks,andindeed,itis$$$$\mathcal{BEAUTY}\mathcal{S}ir!$$

Answered by ajfour last updated on 15/Apr/20

Commented by ajfour last updated on 15/Apr/20

$$sir,something{isn}^{\prime }tjustclear$$$$tomeinthismethod...$$$$Iget4eqs.whileunknowns$$$$arep,q,monly.Please$$$$resolvemydoubtSir.$$

Commented by mr W last updated on 15/Apr/20

$$iwentthroughyourprocessand$$$$couldunderstandallsteps.i{didn}^{\prime }t$$$$seewherethereisthelogicfault.i^{\prime }ll$$$$keepthinkingaboutit.$$

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