Show that the following set of curves intersect orthogonally :

Given:

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

$\& x^{2}-2 y^{2}=4 \ldots(2)$

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

from 2nd curve,

$x^{2}=4+2 y^{2}$

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

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

$\Rightarrow 6 y^{2}=4$

$\Rightarrow y^{2}=\frac{4}{6}$

$\Rightarrow y=\pm \sqrt{\frac{2}{3}}$

Substituting on $y=\pm \sqrt{\frac{2}{3}}$, we get,

$\Rightarrow x^{2}=4+2\left(\pm \sqrt{\frac{2}{3}}\right)^{2}$

$\Rightarrow x^{2}=4+2\left(\frac{2}{3}\right)$

$\Rightarrow x^{2}=4+\frac{4}{3}$

$\Rightarrow x^{2}=\frac{16}{3}$

$\Rightarrow x=\pm \sqrt{\frac{16}{3}}$

$\Rightarrow x=\pm \frac{4}{\sqrt{3}}$

$\therefore$ The point of intersection of two curves $\left(\frac{4}{\sqrt{3}}, \sqrt{\frac{2}{3}}\right) \&\left(-\frac{4}{\sqrt{3}},-\sqrt{\frac{2}{3}}\right)$

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

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

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

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

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

$\Rightarrow x^{2}-2 y^{2}=4$

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

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

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{2 \mathrm{y}} \ldots(4)$

At $\left(\frac{4}{\sqrt{3}}, \sqrt{\frac{2}{3}}\right)$ in equation(3), we get

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\frac{4}{\sqrt{3}}}{4 \times \sqrt{\frac{2}{3}}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\frac{1}{\sqrt{3}}}{\sqrt{\frac{2}{3}}}$

$\Rightarrow m_{1}=\frac{-1}{\sqrt{2}}$

At $\left(\frac{4}{\sqrt{3}}, \sqrt{\frac{2}{3}}\right)$ in equation $(4)$, we get

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{4}{\sqrt{3}}}{\left.2 \times \sqrt{\frac{2}{3}}\right)}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{2}{\sqrt{3}}}{\left.\sqrt{\frac{2}{3}}\right)}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2}{\sqrt{2}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sqrt{2}$

$\Rightarrow m_{2}=1$

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

$\Rightarrow \frac{-1}{\sqrt{2}} \times \sqrt{2}=-1$

$\therefore$ Two curves $x^{2}+4 y^{2}=8 \& x^{2}-2 y^{2}=4$ intersect orthogonally.