Find the angle to intersection of the following curves:

Solution:

Given:

Curves $x^{2}+4 y^{2}=8 \ldots(1)$

$\& x^{2}-2 y^{2}=2 \ldots(2)$]

Solving $(1) \&(2)$, we get,

from 2 nd curve,

$x^{2}=2+2 y^{2}$

Substituting on $x^{2}+4 y^{2}=8$,

$\Rightarrow 2+2 y^{2}+4 y^{2}=8$

$\Rightarrow 6 y^{2}=6$

$\Rightarrow y^{2}=1$

$\Rightarrow y=\pm 1$

Substituting on $y=\pm 1$, we get,

$\Rightarrow x^{2}=2+2(\pm 1)^{2}$

$\Rightarrow x^{2}=4$

$\Rightarrow x=\pm 2$

$\therefore$ The point of intersection of two curves $(2,1) \&(-2,-1)$

Now, Differentiating curves (1) \& (2) w.r.t x, we get

$\Rightarrow x^{2}+4 y^{2}=8$

$\Rightarrow 2 x+8 y \cdot \frac{d y}{d x}=0$

$\Rightarrow 8 y \cdot \frac{d y}{d x}=-2 x$

$\Rightarrow \frac{d y}{d x}=\frac{-x}{4 y} \ldots(3)$

$\Rightarrow x^{2}-2 y^{2}=2$

$\Rightarrow 2 x-4 y \cdot \frac{d y}{d x}=0$

$\Rightarrow x-2 y \cdot \frac{d y}{d x}=0$

$\Rightarrow 4 y \frac{d y}{d x}=x$

$\Rightarrow \frac{d y}{d x}=\frac{x}{2 y} \ldots(4)$

At $(2,1)$ in equation $(3)$, we get

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2}{4 \times 1}$

$\Rightarrow \mathrm{m}_{1}=\frac{-1}{2}$

At $(2,1)$ in equation $(4)$, we get

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{2 \times 1}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1$

$\Rightarrow \mathrm{m}_{2}=1$

when $m_{1}=\frac{-1}{2} \& m_{2}=1$

$\Rightarrow \tan \theta=\left|\frac{\frac{-1}{2}-1}{1+\frac{-1}{2} \times 1}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{-2}{2}}{1-\frac{1}{2}}\right|$

$\Rightarrow \tan \theta=\left|\frac{\frac{-3}{2}}{\frac{1}{2}}\right|$

$\Rightarrow \tan \theta=|-3|$

$\Rightarrow \theta=\tan ^{-1}(3)$

$\Rightarrow \theta \cong 71.56$