Through the vertex $O$ of the parabola $y^2=4x$ variable chords $OP$ and $OQ$ are drawn at right angles. If the variable chord $PQ$ intersects the axis at a fixed point. Find the locus of the middle point of $PQ$.

The correct option is B $y^2=2(x-4)$
Let the equation of the chord $OP$ be $y=mx$ and then,

Equation of the chord $OQ$ will be $=y=-\frac{1}{m}x$ and

$P$ is the point of intersection of $y=mx$ and $y^2=4x$ is $(\frac{4}{m^2},\frac{4}{m})$ and

$Q$ is the point of intersection of $y=-\frac{x}{m}$ and $y^2=4x$ is $(4m^2,-4m)$
Now, the equation of $PQ$ is $y+4m=$ $\frac{\frac{4}{m}+4m}{\frac{4}{m^2}-4m^2}(x-4m^2)$
$\Rightarrow y+4m=\frac{m}{1-m^2}(x-4m^2)$

$\Rightarrow (1-m^2)y+4m-4m^3=mx-4m^3$

$\Rightarrow$ $mx-(1-m^2)y-4m=0$

This line meets $x-$axis, where $y=0$
i.e, $x=4$ $\Rightarrow OL=4$ which is constant as independent of $m$.

Again let $(h,k)$ be the mid-point of $PQ$, then
$h$$=\frac{4m^2+\frac{4}{m^2}}{2}$ and $k$ $=\frac{\frac{4}{m}-4m}{2}$

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

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

Eliminating $m$, we get
$2h=k^2+8$

$y^2=2(x-4)$ is required equation of locus.