Q91 / 200 · Manipal (MET) 2011 · Mathematics
The angle of intersection of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and circle $x^2+y^2=a b$ is
- A$\tan ^{-1}\left(\frac{a+b}{a b}\right)$
- Correct answerB$\tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right)$
- C$\tan ^{-1}\left(\frac{a+b}{\sqrt{a b}}\right)$
- D$\tan ^{-1}\left(\frac{a-b}{a b}\right)$
Solution
Given equation of curves are
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^2+y^2=a b$
$\therefore \quad \frac{a b-y^2}{a^2}+\frac{y^2}{b^2}=1$
$\begin{aligned} & \Rightarrow \quad y^2\left(\frac{a^2-b^2}{a^2 b^2}\right)=\frac{a-b}{a} \\ & \Rightarrow \quad y^2=\frac{a b^2}{a+b}\end{aligned}$
and $\quad x^2=\frac{a^2 b}{a+b}$
$\Rightarrow \quad x=a \sqrt{\frac{b}{a+b}}$
and $\quad y=b \sqrt{\frac{a}{a+b}}$
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 $=-\frac{x}{y}=-\sqrt{\frac{a}{b}}$ $\therefore \quad \theta=\tan ^{-1}\left[\frac{\sqrt{\frac{a}{b}}-\frac{b^2}{a^2} \sqrt{\frac{a}{b}}}{1+\frac{b^2}{a^2} \cdot \frac{a}{b}}\right]$ $\Rightarrow \quad \theta=\tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right)$
Slope of tangent at circle $=-\frac{x}{y}=-\sqrt{\frac{a}{b}}$ $\therefore \quad \theta=\tan ^{-1}\left[\frac{\sqrt{\frac{a}{b}}-\frac{b^2}{a^2} \sqrt{\frac{a}{b}}}{1+\frac{b^2}{a^2} \cdot \frac{a}{b}}\right]$ $\Rightarrow \quad \theta=\tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right)$
Hence, the correct answer is option: 'B'.