Thank you for reporting, we will resolve it shortly
Q.
The locus of the mid point of the line segment joining the point $(4,3)$ and the points on the ellipse $x ^{2}+2 y ^{2}=4$ is an ellipse with eccentricity :
$\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$
Coordinate of $D$ is
$\left(\frac{2 \cos \theta+4}{2}, \frac{\sqrt{2} \sin \theta+3}{2}\right) \equiv(h, k)$
$\frac{2 h-4}{2}=\cos \theta$ ...(i)
$\frac{2 k -3}{\sqrt{2}}=\sin \theta$ ...(ii)
$(i)^{2}+(ii)^{2}$, then we get
$\left(\frac{2 h -4}{2}\right)^{2}+\left(\frac{2 k -3}{\sqrt{2}}\right)^{2}=1$
$\Rightarrow \frac{( x -2)^{2}}{1}+\frac{\left( y -\frac{3}{2}\right)^{2}}{\left(\frac{1}{2}\right)}=1$
$\therefore$ Required eccentricity is
$e =\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$
