The locus of the mid-point of the line segment joining the point $\left(4,3\right)$ and the points on the ellipse $x^2+2y^2=4$ is an ellipse with eccentricity

Given,

$\frac{x^2}{4}+\frac{y^2}{2}=1$

Let coordinate of $D$ $\left(\frac{2\cos \theta +4}{2},\frac{\sqrt{2}\sin \theta +3}{2}\right)\equiv \left(h,k\right)$

So, $\frac{2h-4}{2}=\cos \theta ...\left(1\right)$

And $\frac{2k-3}{\sqrt{2}}=\sin \theta ...\left(2\right)$

$\text{Equation }{\left(1\right)}^2+\text{equation }{\left(2\right)}^2$, then we get

${\left(\frac{2h-4}{2}\right)}^2+{\left(\frac{2k-3}{\sqrt{2}}\right)}^2=1\Rightarrow \frac{{\left(x-2\right)}^2}{1}+\frac{{\left(y-\frac{3}{2}\right)}^2}{\left(\frac{1}{2}\right)}=1$

$\therefore$ Required eccentricity is $e=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$