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

$(-22 / 9,-2 / 9)$

We have,

$2 x^2+y^2=12$                .......(i)

Differentiating w.r.t. $x$, we get

$4 x+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{2 x}{y} \Rightarrow\left(\frac{d y}{d x}\right)_{(2,2)}=-2$

The equation of the normal at $(2,2)$ is

$y-2=\frac{1}{2}(x-2) \Rightarrow x-2 y+2=0$            ......(ii)

Solving (i) and (ii), we obtain that the coordinates of their points of intersection are $(2,2)$ and $(-22 / 9,-2 / 9)$

Hence, the normal to the curve at $(2,2)$ cuts it again at $(-22 / 9,-2 / 9)$.