1. Let the moving point on the ellipse be $P(x,y)$.

   Given ellipse: $x^2+2y^2=4$

2. Let the fixed point be $A(4,3)$.

   If $M(h,k)$ is the midpoint of the segment joining $A(4,3)$ and $P(x,y)$, then by midpoint formula: $h=\frac{x+4}{2},\qquad k=\frac{y+3}{2}$

   So, $x=2h-4,\qquad y=2k-3$

3. Substitute into the ellipse equation: $x^2+2y^2=4$ $\Rightarrow (2h-4)^2+2(2k-3)^2=4$

   Expand: $4(h-2)^2+2(4k^2-12k+9)=4$ $4(h-2)^2+8k^2-24k+18=4$ $4(h-2)^2+8k^2-24k+14=0$

   Now expand completely: $4(h^2-4h+4)+8k^2-24k+14=0$ $4h^2-16h+16+8k^2-24k+14=0$ $4h^2-16h+8k^2-24k+30=0$

   Divide by $2$: $2h^2-8h+4k^2-12k+15=0$

4. Complete the squares: $2(h^2-4h)+4(k^2-3k)+15=0$ $2\big[(h-2)^2-4\big]+4\left[\left(k-\frac32\right)^2-\frac94\right]+15=0$

   $2(h-2)^2-8+4\left(k-\frac32\right)^2-9+15=0$ $2(h-2)^2+4\left(k-\frac32\right)^2=2$

   Divide by $2$: $(h-2)^2+2\left(k-\frac32\right)^2=1$

5. Write in standard form: $\frac{(h-2)^2}{1}+\frac{\left(k-\frac32\right)^2}{\frac12}=1$

   Hence, $a^2=1,\qquad b^2=\frac12$

6. Find the eccentricity: For an ellipse, $e=\sqrt{1-\frac{b^2}{a^2}}$

   So, $e=\sqrt{1-\frac{1/2}{1}}=\sqrt{\frac12}=\frac{1}{\sqrt2}$

7. Option check:

   • A: $\frac{\sqrt3}{2}$ — incorrect
   • B: $\frac{1}{2\sqrt2}$ — incorrect
   • C: $\frac{1}{\sqrt2}$ — correct
   • D: $\frac12$ — incorrect

Therefore, the required eccentricity is $\boxed{\frac{1}{\sqrt2}}$