A) \[{{\tan }^{-1}}\left( \frac{a+b}{ab} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{a-b}{\sqrt{ab}} \right)\]
C) \[{{\tan }^{-1}}\left( \frac{a+b}{\sqrt{ab}} \right)\]
D) \[{{\tan }^{-1}}\left( \frac{a-b}{ab} \right)\]
Correct Answer: B
Solution :
Given equation of curves are \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]and\[{{x}^{2}}+{{y}^{2}}=ab\] \[\therefore \] \[\frac{ab-{{y}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] \[\Rightarrow \] \[{{y}^{2}}\left( \frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}{{b}^{2}}} \right)=\frac{a-b}{a}\] \[\Rightarrow \] \[{{y}^{2}}=\frac{a{{b}^{2}}}{a+b}\] and \[{{x}^{2}}=\frac{{{a}^{2}}b}{a+b}\] \[\Rightarrow \] \[x=a\sqrt{\frac{b}{a+b}}\] and \[y=b\sqrt{\frac{a}{a+b}}\] 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\[=-\frac{x}{y}=-\sqrt{\frac{a}{b}}\] \[\therefore \] \[\theta ={{\tan }^{-1}}\left[ \frac{\sqrt{\frac{a}{b}}-\frac{{{b}^{2}}}{{{a}^{2}}}\sqrt{\frac{a}{b}}}{1+\frac{{{b}^{2}}}{{{a}^{2}}}\cdot \frac{a}{b}} \right]\] \[\Rightarrow \] \[\theta ={{\tan }^{-1}}\left( \frac{a-b}{\sqrt{ab}} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
