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Question Number 28464 by ajfour last updated on 26/Jan/18

Answered by mrW2 last updated on 26/Jan/18

$$eqn.ofellipse:$$$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$$$\frac{2x}{a^2}+\frac{2y}{b^2}\times \frac{dy}{dx}=0$$$$\Rightarrow \frac{dy}{dx}=-\frac{b^2}{a^2}\times \frac{x}{y}=inclinationoftangent$$$$\Rightarrow \tan \phi =inclinationofnormal=\frac{a^2}{b^2}\times \frac{y}{x}$$$$\frac{y}{x}=\tan \varphi$$$$\Rightarrow \tan \phi =\frac{a^2}{b^2}\tan \varphi$$$$\theta =\phi -\varphi$$$$\Rightarrow \tan \theta =\frac{\frac{a^2}{b^2}\tan \varphi -\tan \varphi }{1+\frac{a^2}{b^2}{\tan }^2\varphi }=\frac{(1-\frac{b^2}{a^2})\tan \varphi }{\frac{b^2}{a^2}+{\tan }^2\varphi }=\frac{(1-k^2)\tan \varphi }{k^2+{\tan }^2\varphi }$$$$withk=\frac{b}{a}$$$$\Rightarrow \tan \theta =\frac{(1-k^2)\tan \varphi }{k^2+{\tan }^2\varphi }=\frac{1-k^2}{\frac{k^2}{\tan \varphi }+\tan \varphi }⩽\frac{1-k^2}{2\sqrt{\frac{k^2}{\tan \varphi }\times \tan \varphi }}=\frac{1-k^2}{2k}$$$$i.e.\tan {\theta }_{max}=\frac{1-k^2}{2k}=\frac{a^2-b^2}{2ab}$$$$or{\theta }_{max}={\tan }^{-1}\frac{1-k^2}{2k}={\tan }^{-1}\frac{a^2-b^2}{2ab}$$$$at\frac{k^2}{\tan \varphi }=\tan \varphi or\varphi =\pm {\tan }^{-1}k=\pm {\tan }^{-1}\frac{b}{a}$$

Commented by ajfour last updated on 26/Jan/18

$$great,suggestsashortermethod$$$$hopefully!$$

Commented by ajfour last updated on 26/Jan/18

$$ThanksSir.$$

Commented by mrW2 last updated on 26/Jan/18

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