The normal to the curve $2{\mathrm{x}}^2+{\mathrm{y}}^2=12$ at the point (2, 2) cuts the curve again at 

$2{\mathrm{x}}^2+{\mathrm{y}}^2=12 \mathrm{or} \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{2\mathrm{x}}{\mathrm{y}}$

Slope of normal at point $\mathrm{A}\left(2,2\right) \mathrm{is} \frac{1}{2}$

Also. point $\mathrm{B}\left(-\frac{22}{9},-\frac{2}{9}\right)$ lies on the curve and slope of AB is

$\frac{2-\left(-\frac{2}{9}\right)}{2-\left(-\frac{22}{9}\right)}=\frac{1}{2}.$

Hence, the normal meets the curve again at point  $\left(-\frac{22}{9},-\frac{2}{9}\right)$