Question

An angle of intersection of the curves, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^2+y^2=ab,$ $a>b$, is

• A. ${\tan }^{-1}\left(\frac{a+b}{\sqrt{ab}}\right)$
• B. ${\tan }^{-1}\left(\frac{a-b}{2\sqrt{ab}}\right)$
• C. ${\tan }^{-1}\left(\frac{a-b}{\sqrt{ab}}\right)$
• D. ${\tan }^{-1}\left(2\sqrt{ab}\right)$

Answer 1

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

Let point of intersection be $(x_1,y_1)$
$x_1^2+y_1^2-ab=0$ and
$b^2{x}_1^2+a^2{y}_1^2-a^2{b}^2=0$
We get
$x_1^2=\frac{a^{2}b}{a+b},y_1^2=\frac{a{b}^2}{a+b}$
$\Rightarrow x_1=a\sqrt{\frac{b}{a+b}},y_1=b\sqrt{\frac{a}{a+b}}\cdots \left(1\right)$
Tangent to ellipse: $\frac{x{x}_1}{a^2}+\frac{y{y}_1}{b^2}=1$
Slope $=m_1=-\frac{b^2{x}_1}{a^2{y}_1}$
Tangent to circle: $x{x}_1+y{y}_1=ab$
Slope $=m_2=-\frac{x_1}{y_1}$
Let angle between both curves be $\theta$
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\frac{-b^2{x}_1}{a^2{y}_1}+\frac{x_1}{y_1}}{1+\frac{b^2{x}_1}{a^2{y}_1}\cdot \frac{x_1}{y_1}}\right|=\frac{x_1{y}_1(a^2-b^2)}{b^2{x}_1^2+a^2{y}_1^2}=\frac{x_1{y}_1(a^2-b^2)}{a^2{b}^2}$
$=\frac{ab\sqrt{ab}}{a+b}\cdot \frac{(a^2-b^2)}{a^2{b}^2}$ using $\left(1\right)$
$\Rightarrow \tan \theta =\frac{a-b}{\sqrt{ab}}$