Question

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

Answer 1

Answer: (B)

Given equation of curves are
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and $x^{2}+y^{2}=a b$
$\therefore \frac{a b-y^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
$\Rightarrow y^{2}\left(\frac{a^{2}-b^{2}}{a^{2} b^{2}}\right)=\frac{a-b}{a}$
$\Rightarrow y^{2}=\frac{a b^{2}}{a+ b}$
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 $=-\frac{x}{y}=-\sqrt{\frac{a}{b}}$
$\therefore \theta=\tan ^{-1}\left[\frac{\sqrt{\frac{a}{b}}-\frac{b^{2}}{a^{2}} \sqrt{\frac{a}{b}}}{1+\frac{b^{2}}{a^{2}} \cdot \frac{a}{b}}\right]$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right)$