Question

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. ${\tan }^{-1}\left(\frac{a-b}{\sqrt{ab}}\right)$
• C. ${\tan }^{-1}\left(\frac{a+b}{\sqrt{ab}}\right)$
• D. ${\tan }^{-1}\left(\frac{a-b}{ab}\right)$

Answer 1

Answer: (B)

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 $=-\frac{x}{y}=-\sqrt{\frac{a}{b}}$
$\therefore \theta ={\tan }^{-1}\left[\frac{\sqrt{\frac{a}{b}}-\frac{b^2}{a^2}\sqrt{\frac{a}{b}}}{1+\frac{b^2}{a^2}\cdot \frac{a}{b}}\right]$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{a-b}{\sqrt{ab}}\right)$