Shortest distance always lies along the common normal
$\Rightarrow$ Slope of common normal is $1$
Now, the equation of the normal at the point $\left(\mathrm{acos}\theta ,\mathrm{bsin}\theta \right) \mathrm{on} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$\frac{ax}{\mathrm{c}\mathrm{o}\mathrm{s}\theta }-\frac{by}{\mathrm{s}\mathrm{i}\mathrm{n}\theta }=a^2-b^2,$ where slope is $\frac{a}{b}\tan \theta$
$\Rightarrow \frac{\sqrt{6}}{\sqrt{3}}\tan \theta =1\Rightarrow \tan \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \sin \theta =\frac{1}{\sqrt{3}}$ and $\cos \theta =\frac{\sqrt{2}}{\sqrt{3}}$
or $\sin \theta =-\frac{1}{\sqrt{3}}, \cos \theta =-\frac{\sqrt{2}}{\sqrt{3}}$ (rejected because it gives farthest point)
$\Rightarrow$ Nearest point is $\left(\sqrt{6}\left(\frac{\sqrt{2}}{\sqrt{3}}\right),\sqrt{3}\left(\frac{1}{\sqrt{3}}\right)\right)=\left(\mathrm{2,1}\right)$