Given ellipse
$2 x^2+y^2=2 \text { or } x^2+\frac{y^2}{2}=1$
Here, $\quad a^2=1, b^2=2$
Equation of tangent
$x+2 y+k=0 \text { or } 2 y=-x-k$
or $\quad y=-\frac{x}{2}-\frac{k}{2}$
Here,
$c=\frac{-k}{2}, m=-\frac{1}{2}$
From the condition of tangent
$c= \pm \sqrt{a^2 m^2+b^2} \Rightarrow \frac{-k}{2}= \pm \sqrt{(1)\left(\frac{1}{4}\right)+2}$
$\frac{-k}{2}= \pm \sqrt{\frac{9}{4}} \Rightarrow \frac{-k}{2}= \pm \frac{3}{2}$ or $k= \pm 3$
But $k>0$, so, $k=3$
So, equation of tangent, $y=-\frac{x}{2}-\frac{3}{2}$ and point is $\left(\frac{1}{\sqrt{2}}, 1\right)$
The equation of the normal to the ellipse at point $\left(x_1, y_1\right)$ is given by
$\frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2 \Rightarrow \frac{1 \times x}{\frac{1}{\sqrt{2}}}-\frac{2 \times y}{1}=1-2$
$\sqrt{2} x-2 y=-1$ or $\sqrt{2} x-2 y+1=0$