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

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

We have,

$C_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$           .....(i)

and $C_2: x^2+y^2=a b$        .....(ii)

Let $P\left(x_1, y_1\right)$ be the point of intersection of these curves. Then, at point P, we have

$\left(\frac{d y}{d x}\right)_{C_1}=-\frac{b^2 x_1}{a^2 y_1}$ and, $\left(\frac{d y}{d x}\right)_{C_2}=-\frac{x_1}{y_1}$

Let $\theta$ be the angle of intersection of (i) and (ii) at point P. Then,

$\tan \theta=\left|\frac{\left(\frac{d y}{d x}\right)_{C_1}-\left(\frac{d y}{d x}\right)_{C_2}}{1+\left(\frac{d y}{d x}\right)_{C_1}\left(\frac{d y}{d x}\right)_{C_2}}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{-b^2 x_1}{a^2 y_1}+\frac{x_1}{y_1}}{1+\frac{b^2 x_1{ }^2}{a^2 y_1{ }^2}}\right|$

$\Rightarrow \tan \theta=\frac{\left(a^2-b^2\right) x_1 y_1}{a^2 y_1{ }^2+b^2 x_1{ }^2}$

$\Rightarrow \tan \theta=\frac{a^2-b^2}{a^2 b^2} \times \sqrt{\frac{a^3 b^3}{(a+b)^2}}$

Upon solving (i) and (ii)

we get $x=\sqrt{\frac{a^2b}{a+b}},y=\sqrt{\frac{ab^2}{a+b}}$

$\Rightarrow \tan \theta=\left(\frac{a-b}{\sqrt{a b}}\right)$