P(h, k) is a fixed point in the plane of the parabola $y^2=4ax.$ Through $P$ a variable secant is drawn to cut the parabola in Q and R. T is a point on QR such that $PQ.PR=P{T}^2$ then the locus of T is

$\frac{x-h}{\cos \theta }=\frac{y-k}{\sin \theta }=r$

$(x,y)=(h+r\cos \theta ,k+r\sin \theta )$

If that point lies on the parabola ${(k+r\sin \theta )}^2=4ach+4\cos \theta$

$r^2{\sin }^2\theta +2r(k\sin \theta -2a\cos \theta )+(k^2-4ah)=0$

$r_1{r}_2=\frac{k^2-4ah}{{\sin }^2\theta }$

$\left(PQ\right)\left(PR\right)=\frac{k^2-4ah}{{\sin }^2\theta }$

Let T be $(\alpha ,\beta )$

Slope of $PT=\frac{\beta -k}{\alpha -h}=\tan \theta$

${\sin }^2\theta =\frac{{(\beta -k)}^2}{{(\alpha -h)}^2+{(\beta -k)}^2}=\frac{{(\beta -k)}^2}{P{T}^2}$

$\left(PQ\right)\left(PR\right)=\frac{k^2-4ah}{{(\beta -k)}^2}P{T}^2\Rightarrow k^2-4ah={(\beta -k)}^2\Rightarrow {(y-k)}^2=k^2-4ah\Rightarrow y=k\pm \sqrt{k^2-4ah}$

(h, k) lies outside the parabola $\Rightarrow k^2-4ah>0\Rightarrow y=k+\sqrt{k^2-4ah},y=k-\sqrt{k^2-4ah}$

(h, k) lies on the parabola $\Rightarrow k^2=4ah\Rightarrow y=k$

A line parallel to x–axis