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) \[{{\tan }^{-1}}\left( \frac{a-b}{\sqrt{ab}} \right)\]
Correct Answer: D
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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