A = (–2, 0) and P is a point on the parabola y² = 8x. If Q bisects $\overline{AP}$ and the locus of Q is a parabola then its focus is

$\begin{array}{l}Given parabola y^2=8x\\ \Rightarrow 4\mathrm{a}=8\Rightarrow a=2\\ \Rightarrow \mathrm{A}=(-2,0);\mathrm{P}=\left(2{\mathrm{t}}^2,4\mathrm{t}\right)\\ \mathrm{Given} \mathrm{Q}=\mathrm{Mid} \mathrm{point}\text{ of AP}\\ (\mathrm{x},\mathrm{y})=\left(\frac{-2+2{\mathrm{t}}^2}{2},\frac{4\mathrm{t}}{2}\right)\\ \Rightarrow (\mathrm{x},\mathrm{y})=\left({\mathrm{t}}^2-1,2\mathrm{t}\right)\\ \Rightarrow \mathrm{x}={\mathrm{t}}^2-1,\mathrm{y}=2\mathrm{t}\\ \Rightarrow {\mathrm{y}}^2=4{\mathrm{t}}^2\Rightarrow {\mathrm{y}}^2=4(\mathrm{x}+1)\\ focus=(h+a,k)=(0,0)\end{array}$