The number of integral values of a for which three distinct chords of the ellipse $\frac{x^2}{2a^2}+\frac{y^2}{a^2}=1$  passing through the point  $(20a,-\frac{a^2}{2})$ are bisected by the parabola  $y^2=4ax$  is

Any point on the parabola $y^2=4ax$  is  $(a^2,2at)$
Equation of the chord of the ellipse  $\frac{x^2}{2a^2}+\frac{y^2}{a^2}=1$
Whose mid-point is (at $,2at)$   is  $T=S_1$
 $\Rightarrow x\frac{a{t}^2}{2a^2}+y\frac{2at}{a^2}=\frac{a^2{t}^4}{2a^2}+\frac{4a^2{t}^2}{a^2}\Rightarrow tx+4y+a{t}^3+8at$
As it passes through  $(20a,-\frac{a^2}{2})\Rightarrow 20at+4(-\frac{a^2}{2})=a{t}^3+8at$
 $\Rightarrow a{t}^3-12at+2a^2=0\Rightarrow t^3-12t+2a=0(a\ne 0)$
Now three should be 3 distinct values of t for three distinct chords
Let  $f\left(t\right)=t^3-12t+2a;f^{\text{'}}\left(t\right)=3t^2-12=0\Rightarrow t=\pm 2$
So  $f\left(2\right)f(-2)<0\Rightarrow (a+8)(a-8)<0$
$\therefore 14$  values of a are possible  $(a\ne 0)$