Locus of vertex of moving parabola

Question

The parabolas $y^2=-12x$ and $y^2=12x$ touch each other at $(0,0)$. The parabola $y^2=-12x$ starts rolling on the parabola $y^2=12x$ (without slipping). Find the locus of the vertex of the moving parabola.

My Attempt:

The two parabolas are always going to have a common tangent at the point of contact. Is it possible to write an equation of a family of curves which touch this variable tangent?

https://math.stackexchange.com/questions/63025/why-does-the-focus-of-a-rolling-parabola-trace-a-catenary — lab bhattacharjee, Sep 02 '17 at 06:47
https://math.stackexchange.com/questions/32629/a-circle-rolls-along-a-parabola — lab bhattacharjee, Sep 02 '17 at 06:58

score 4 · Answer 1 · answered Sep 04 '17 at 03:40

As noted here (en Français) and here, the roulette traced by a fixed point on a curve rolling on a copy of this curve is the same as the orthotomic (effectively a scaled pedal curve) with respect to that fixed point.

For the particular case of the parabola, the orthotomic (pedal) with respect to the vertex is called the cissoid of Diocles.

cissoid

(image taken from here)

score 3 · Accepted Answer · answered Sep 02 '17 at 15:36

Let $P(3t^2,6t)$ be any point on the parabola $y^2=12x$. The equation of tangent to the parabola at $P$ is

$ty=x+3t^2$ .

Now the vertex of the parabola (say $(h,k)$) which has rolled will be image of $(0,0)$ with respect to tangent line $ty=x+3t^2$.

$\frac{h-0}{1}=\frac{k-0}{-t}=-\frac{2(0-t(0)+3t^2)}{1+t^2}$

Thus, $t=-\frac{k}{h}$

Eliminating $t$ , we get

$h(h^2+k^2)+6k^2=0$

So required locus should be

$x(x^2+y^2)+6y^2=0$

Locus of vertex of moving parabola

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