Let the midpoint of the required chord be $M(h,k)$.

We need the locus of the midpoint of chords of the hyperbola $x^2-y^2=4$ such that the chord touches the parabola $y^2=8x.$

Equation of chord of the hyperbola with given midpoint $(h,k)$.

For the hyperbola $S\equiv x^2-y^2-4=0,$ the chord whose midpoint is $(h,k)$ is given by the midpoint form: $T=S_1,$ where $T=xx_1-yy_1-4, \qquad S_1=x_1^2-y_1^2-4.$

Putting $(x_1,y_1)=(h,k)$, $xh-yk-4=h^2-k^2-4.$ Hence, $xh-yk=h^2-k^2.$

So the chord is hx-ky=h^2-k^2. \tag{1}

Condition that this chord touches the parabola $y^2=8x$.

Write the line (1) as $hx-ky-(h^2-k^2)=0.$

If $h\neq 0$, then $x=\frac{k}{h}y+\frac{h^2-k^2}{h}.$

Substitute into the parabola equation: $y^2=8x=8\left(\frac{k}{h}y+\frac{h^2-k^2}{h}\right).$

Rearranging, $hy^2-8ky-8(h^2-k^2)=0.$

For tangency, discriminant must be zero: $(-8k)^2-4(h)(-8(h^2-k^2))=0.$

$64k^2+32h(h^2-k^2)=0.$

Divide by $32$: $2k^2+h(h^2-k^2)=0.$

$h^3-hk^2+2k^2=0.$

$h^3+k^2(2-h)=0.$

Therefore, $k^2(h-2)=h^3.$

Hence the locus of $(h,k)$ is $y^2(x-2)=x^3.$

Match with options.

This is exactly Option C: $y^2(x-2)=x^3.$

Comparison with stored correct answer.

Stored correct answer: C.

Our derived answer also gives C. So they agree.