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

 $\frac{x^2}{4}+\frac{y^2}{2}=1$
Let parametric point on ellipse is  $(2\cos \theta ,\sqrt{2}\sin \theta )$
Let mid point R
 
Coordinate of R is
$(\frac{2\cos \theta +4}{2}.\frac{\sqrt{2}\sin \theta +3}{2})\equiv (h,k)$ 
$\frac{2h-4}{2}=\cos \theta$                                   -------(i)
$\frac{2k-3}{\sqrt{2}}=\sin \theta$                                         ----------------------(ii)
On squaring equations (i) and (ii), then w get
 ${\left(\frac{2h-4}{2}\right)}^2+{\left(\frac{2k-3}{\sqrt{2}}\right)}^2=1\Rightarrow \frac{{(x-2)}^2}{1}+\frac{{(y-\frac{3}{2})}^2}{\left(\frac{1}{2}\right)}=1$