Question

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

• A. $(0,0)$
• B. $(1,1)$
• C. $(5,0)$
• D. $(4,0)$

Answer 1

The correct option is A $(0,0)$

Let P be $(2t^2,4t)$ and Q be $(h,k)$

Also given $A=(-2,0)$

Now since $Q$ is midpoint of $AP$,

$h=\frac{2t^2-2}{2}=t^2-1$

and $k=\frac{4t+0}{2}=2t$

Now eliminating $t$ from above two,

$h=\frac{k^2}{2}-1\Rightarrow k^2=4(h+1)$

So the locus of $Q(h,k)$ is $y^2=4(x+1)$,

which clearly a parabola having focus $(0,0)$