# Locus of a point from where product of length of tangents to parabola equals product of its latus rectum and focal distance of the point

If the product of the lengths of the tangents drawn from a point $P$ to the parabola $y^2=4ax$ is equal to the product of the focal distance of the point P and the latus rectum. Prove that the locus of $P$ is the parabola $y^2=4a(a+x)$

My work. I assumed $$\frac{x-s}{p} =\frac{y-t}{q}=r,$$ where $p^2 +q^2=1$, as the equation of tangent of parabola through point $P=(s, t)$ and assumed that this tangent will touch given parabola at a distance $r$ from $(s,t)$. Then $(s+pr,t+qr)$ must lie on parabola. Hence $$(t+qr)^2=4a(s+pr)$$ or,
$$q^2r^2-2r(2ap-qt)+(t^2-4as)=0\tag{1}$$ The two roots are the distance between $(s, t)$ and the two points on parabola where the pair of tangents from $(s, t)$ on it touches it. So, product of length of tangents from $(s, t)$ is the product of the roots of $(1)$, that is $$\frac{t^2-4as}{q^2}.$$ From this step how to proceed?

In my opinion it is better to start from the point of tangency. The tangents to the parabola $x=y^2/(4a)$ at $y_1=2s$ and $y_2=2t$ have equations $$x=\frac{s}{a}+\frac{s}{a}\left(y-2s\right)\;\;,\;\;x=\frac{t}{a}+\frac{t}{a}\left(y-2t\right).$$ Their intersection is $$P= \left(\frac{st}{a},s+t\right).$$ Hence the locus is $y^2=4a(a+x)$ if and only if $$\left(s+t\right)^2=4a\left(a+\frac{st}{a}\right)\Leftrightarrow (s-t)^2=4a^2.$$ The square of the product of the lengths of the tangents drawn from a point P to the parabola is $$\left(\left(\frac{st}{a}-\frac{s^2}{a}\right)^2+\left(s+t-2s\right)^2\right)\cdot \left(\left(\frac{st}{a}-\frac{t^2}{a}\right)^2+\left(s+t-2t\right)^2\right)$$ that is $$\frac{(s-t)^4 (s^2+a^2)(t^2+a^2)}{a^4}\tag{1}.$$ Moreover the focus has coordinates $(a,0)$ and the latus rectum is $4a$. Hence, the square of the product of the focal distance of the point P and the latus rectum is $$\left(\left(\frac{st}{a}-a\right)^2+\left(s+t-0\right)^2\right)\cdot (4a)^2,$$ that is $$16(s^2+a^2)(t^2+a^2) \tag{2}.$$ Finally, by equating (1) and (2), we obtain $(s-t)^4=16a^4$, that is $(s-t)^2=4a^2$ and we are done.