Question

The intersection angle of between the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the circle $x^2+y^2=ab$ is:

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

Answer 1

The correct option is D ${\tan }^{-1}\frac{(b-a)}{\sqrt{ab}}$

Let us find the point of intersection of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ...(1)

and circle $x^2+y^2=ab$ ...(2)

We have, $\frac{x^2}{a^2}+\frac{1}{b^2}(ab-x^2)=1\Rightarrow x^2(\frac{1}{a^2}-\frac{1}{b^2})=1-\frac{a}{b}=\frac{b-a}{b}$

$\Rightarrow x^2\frac{(b^2-a^2)}{a^2{b}^2}=\frac{b-a}{b}\Rightarrow x^2=\frac{a^{2}b}{a+b}$ ...(3)

and $y^2=ab-x^2\Rightarrow y^2=ab-\frac{a^{2}b}{a+b}$ (Using (1) and (3))

$\Rightarrow y^2=\frac{ab(a+b-a)}{a+b}=\frac{a{b}^2}{a+b}$

$\therefore$ Intersection point is $(\sqrt{\frac{a^{2}b}{a+b}},\sqrt{\frac{a{b}^2}{a+b}})$

The slope of tangent to ellipse at any point $(x_1,y_1)$ is

$m_1=-\frac{b^2{x}_1}{a^2{y}_1}=-\frac{b^2}{a^2}\frac{\sqrt{\frac{a^{2}b}{a+b}}}{\sqrt{\frac{a{b}^2}{a+b}}}=-\frac{b^2}{a^2}\sqrt{\frac{a}{b}}$

and slope of tangent to the circle is

$m_2=-\frac{x_1}{y_1}=-\sqrt{\frac{a}{b}}$

$\therefore \theta ={\tan }^{-1}\left(\frac{-\sqrt{\frac{a}{b}}+\frac{b^2}{a^2}\sqrt{\frac{a}{b}}}{1+\frac{b^2}{a^2}.\frac{a}{b}}\right)={\tan }^{-1}\frac{(b-a)}{\sqrt{ab}}$