Question

Let $P$ be the point $(1,0)$ and $Q$ a point on the parabola $y^2=8x$, then find the locus of the mid-point of $PQ$

• A. $y^2+2x+2=0$
• B. $y^2-4x+2=0$
• C. $y^2-2x-2=0$
• D. $y^2+4x+2=0$

Answer 1

Answer: (B)

$y^2-4x+2=0$

Given $P\equiv (1,0)$
let $Q\equiv (2t^2,4t),$ (Since $a=2$)
Let mid point of $PQ$ be $(h,k)$
$\Rightarrow h=\frac{2t^2+1}{2},k=\frac{4t+0}{2}=2t$
Eliminating $t$ we get $k^2-4h+2=0$
Hence locus of mid point of $PQ$ is, $y^2-4x+2=0$