A = (–2, 0) and P is a point on the parabola $y^2 = 8x$. If Q bisects $\overline{AP}$ and the locus of Q is a parabola then its focus is

$4a=8;\Rightarrow a=2,\mathrm{A}=(-2,0)$.

$\mathrm{P}=\left(2t^2,4t\right);\mathrm{Q}=\text{ M.P of A.P, }$

$(x,y)=\left(\frac{-2+2t^2}{2},\frac{4t}{2}\right)\Rightarrow (x,y)=\left(t^2-1,2t\right)\text{, }$

$x=t^2-1,y=2t,y^2=4t^2,y^2=4(x+1)$

vertex(h,k)=(-1,0)

Focus=(h+a,k)=(0,0)