Find latus rectum of locus of centroid of equilateral triangle inscribed in parabola x^2 = 36y? This is not a homework help question, I found a problem when I was doing self-study , but I am getting stuck. Please tell me if my method is right, or if there are other steps I need to take.
Since the parabola is in the form $x^2 = 4ay$ where $a=9$, I deduced that the vertices of the triangle will be $A(9t_1^2,18t_1)$, $B(9t_2^2,18t_2)$ and $C(9t_3^3,18t_3)$. The $x$ and $y$ coordinates of centroid $h$ and $k$ will thus be given by the sum of coordinates of the vertices
$h = 6(t_1+t_2+t_3)$ and $y = 3(t_1^2+t_2^2+t_3^2)$.
However I do not know how to find out if this is a conic section.
I also wondered if I should use the fact that the centroid divides the median in the ratio 2:1, and thus got $h = 3t_1^2+6(t_2+t_3)$ and $k = 6t_1 + 3(t_2^2+t_3^2)$
However none of these incorporate the fact that the triangle is an equilateral triangle. I do not know how to proceed or what method to take. Would someone please help me? The answer is that the length of the latus rectum is 4.
 A: Given parabola $\mathcal{P} : 4ay = x^2$ with parametrization $\mathbb{R} \ni t \mapsto \rho(t) = (2at,at^2) \in \mathbb{R}^2$.
Let $v_1, v_2, v_3$ be any $3$ points on $\mathcal{P}$ whose parameters $t_1, t_2, t_3$ satisfy $t_1 < t_2 < t_3$.
Consider the triangle formed using these vertices. Let 


*

*$\theta_1$ be the angle between the side $v_1v_2$ with $x$-axis.

*$\theta_2$ be the angle between the side $v_2v_3$ with $x$-axis


It is easy to see
$$
\tan\theta_1 = \frac{at_2^2 - at_1^2}{2at_2 - 2at_1} = \frac{t_2+t_1}{2}\quad\text{ and }\quad
\tan\theta_2 = \frac{at_3^2 - at_2^2}{2at_3 - 2at_2} = \frac{t_3+t_2}{2}
$$
In order for $v_1, v_2, v_3$ to form an equilateral triangle, the exterior
angle at $v_2$ has to be $120^\circ$.
This means
$$\frac{\tan\theta_2 - \tan\theta_1}{1 + \tan\theta_2\tan\theta_1}
= \tan(\theta_2-\theta_1) = \tan 120^\circ = -\sqrt{3}$$ 
Plugging expression of $\tan\theta_1$, $\tan\theta_2$ into above and simplify.
Do the same thing to other two vertices, we obtain:
$$
\begin{align}
4 + (t_3+t_2)(t_2+t_1) + \frac{2}{\sqrt{3}}(t_3 - t_1) = 0\\
4 + (t_1+t_3)(t_3+t_2) + \frac{2}{\sqrt{3}}(t_1 - t_2) = 0\\
4 + (t_2+t_1)(t_1+t_3) + \frac{2}{\sqrt{3}}(t_2 - t_3) = 0
\end{align}
$$
Summing these equations, we obtain
$$\begin{align}
 &(t_1+t_2)(t_2+t_3) + (t_2+t_3)(t_3+t_1)+(t_3+t_1)(t_1+t_2) + 12 = 0\\
\iff & t_1^2+t_2^2+t_3^2 + 3(t_1t_2+t_2t_3+t_3t_1) + 12 = 0\\
\iff & t_1^2+t_2^2+t_3^2 = 3(t_1+t_2+t_3)^2 + 24
\end{align}
$$
The centroid of the triangle is located at
$(x_c,y_c) = \left(\frac{2a}{3}(t_1+t_2+t_3),\frac{a}{3}(t_1^2+t_2^2+t_3^2)\right)$. Above equality implies $(x_c,y_c)$ is lying on the curve
$$\frac{3y_c}{a} = 3\left(\frac{3x_c}{2a}\right)^2 + 24
\quad\iff\quad 4ay_c = 9 x_c^2 + 32a^2
$$
For $a = 9$, this becomes $4 y_c = x_c^2 + 288$, a parabola with latus rectum $4$.
There are other ways to show the locus is a parabola.
For an alternate approach using complex numbers, see
my answer to a related
question about the parabola $y = x^2$.
