• # question_answer The angle of intersection of ellipse $\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$ and circle ${{x}^{2}}+{{y}^{2}}=ab$, is A)            ${{\tan }^{-1}}\left( \frac{a-b}{ab} \right)$                               B)            ${{\tan }^{-1}}\left( \frac{a+b}{ab} \right)$ C)            ${{\tan }^{-1}}\left( \frac{a+b}{\sqrt{ab}} \right)$                    D)            ${{\tan }^{-1}}\left( \frac{a-b}{\sqrt{ab}} \right)$

Solution :

$\frac{ab-{{y}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1$or ${{y}^{2}}\left( \frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}{{b}^{2}}} \right)=\frac{a-b}{a}$            or${{y}^{2}}=\left( \frac{a{{b}^{2}}}{a+b} \right)$and ${{x}^{2}}=\left( \frac{{{a}^{2}}b}{a+b} \right)\,\,\Rightarrow \,(x,\,y)=\left( a\sqrt{\frac{b}{a+b}},b\sqrt{\frac{a}{a+b}} \right)$            Slope of tangent at ellipse $=\frac{-{{b}^{2}}x}{{{a}^{2}}y}=\frac{-{{b}^{2}}}{{{a}^{2}}}\sqrt{\frac{a}{b}}$            Slope of tangent at circle $=-\frac{x}{y}=-\sqrt{\frac{a}{b}}$            $\therefore \theta ={{\tan }^{-1}}\left[ \frac{-\sqrt{\frac{a}{b}}+\frac{{{b}^{2}}}{{{a}^{2}}}\sqrt{\frac{a}{b}}}{1+\frac{{{b}^{2}}}{{{a}^{2}}}.\frac{a}{b}} \right]$  i.e., $\theta ={{\tan }^{-1}}\left[ \frac{a-b}{\sqrt{ab}} \right]$.                    Note : Students should remember this question as a formula.

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