# An angle of intersection of the curves,

Question:

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

and $x^{2}+y^{2}=a b, a>b$, is :

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

2. $\tan ^{-1}\left(\frac{a-b}{2 \sqrt{a b}}\right)$

3. $\tan ^{-1}\left(\frac{a-b}{\sqrt{a b}}\right)$

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

Correct Option: , 3

Solution:

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, x^{2}+y^{2}=a b$

$\frac{2 x_{1}}{a^{2}}+\frac{2 y_{1} y^{\prime}}{b^{2}}=0$

$\Rightarrow \mathrm{y}_{1}{ }^{\prime}=\frac{-\mathrm{x}_{1}}{\mathrm{a}^{2}} \frac{\mathrm{b}^{2}}{\mathrm{y}_{1}}$    ..........(1)

$\therefore 2 \mathrm{x}_{1}+2 \mathrm{y}_{1} \mathrm{y}^{\prime}=0$

$\Rightarrow \mathrm{y}_{2}^{\prime}=\frac{-\mathrm{x}_{1}}{\mathrm{y}_{1}}$           ............(2)

Here $\left(x_{1} y_{1}\right)$ is point of intersection of both curves

$\therefore x_{1}^{2}=\frac{a^{2} b}{a+b}, y_{1}^{2}=\frac{a b^{2}}{a+b}$

$\therefore \tan \theta=\left|\frac{\mathrm{y}_{1}{ }^{\prime}-\mathrm{y}_{2}{ }^{\prime}}{1+\mathrm{y}_{1}{ }^{\prime} \mathrm{y}_{2}{ }^{\prime}}\right|=\left|\frac{\frac{-\mathrm{x}_{1} \mathrm{~b}^{2}}{\mathrm{a}^{2} \mathrm{y}_{1}}+\frac{\mathrm{x}_{1}}{\mathrm{y}_{1}}}{1+\frac{\mathrm{x}_{1}{ }^{2} \mathrm{~b}^{2}}{\mathrm{a}^{2} \mathrm{y}_{1}^{2}}}\right|$

$\tan \theta=\left|\frac{-b^{2} x_{1} y_{1}+a^{2} x_{1} y_{1}}{a^{2} y_{1}^{2}+b^{2} x_{1}^{2}}\right|$

$\tan \theta=\left|\frac{\mathrm{a}-\mathrm{b}}{\sqrt{\mathrm{ab}}}\right|$