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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?

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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.

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  • $\begingroup$ @Samya Any further doubt? Did you make any progress with your approach? $\endgroup$
    – Robert Z
    Oct 23 '17 at 11:23

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