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. ${\tan }^{-1}(2\sqrt{ab})$

Answer 1

Answer: (C)

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,x^2+y^2=ab$
$\frac{2x_1}{a^2}+\frac{2y_1{y}^{\prime }}{b^2}=0$
$\Rightarrow y_1^{\prime }=\frac{-x_1}{a^2}\frac{b^2}{y_1}...$(1)
$\therefore 2x_1+2y_1{y}^{\prime }=0$
$\Rightarrow y_2^{\prime }=\frac{-x_1}{y_1}...$(2)
Here $\left(x_1{y}_1\right)$ is point of intersection of both curves
$\therefore x_1^2=\frac{a^{2}b}{a+b},y_1^2=\frac{a{b}^2}{a+b}$
$\therefore \tan \theta =\left|\frac{y_1^{\prime }-y_2^{\prime }}{1+y_1^{\prime }{y}_2^{\prime }}\right|=\left|\frac{\frac{-x_1{b}^2}{a^2{y}_1}+\frac{x_1}{y_1}}{1+\frac{x_1^2{b}^2}{a^2{y}_1^2}}\right|$
$\tan \theta =\left|\frac{-b^2{x}_1{y}_1+a^2{x}_1{y}_1}{a^2{y}_1^2+b^2{x}_1^2}\right|$
$\tan \theta =\left|\frac{a-b}{\sqrt{ab}}\right|$