<s>[INST] Locus problem on circle and parabola 
Circles are drawn through the vertex of a parabola $y^2 = 4ax$ to cut the parabola orthogonally at the other point. Find the locus of the centers of the circles.

What I did:
The tangent must surely be the diameter of the circle. Also, the line through the vertex perpendicular to the line joining the vertex to the point on parabola must intersect the circle at the other end of the diameter i.e. where the tangent meets the circle again. The center of the circle must be their midpoint

So, I take a parametric point $P(at^2, 2at)$ on the parabola. The equation of the line perpendicular to $VP$ going through $V$ is $2y = -tx$. It meets the tangent $ty = x + at^2$ at $G$ which is the other diametric end. The coordinates of $G$ are $\Large(\frac{-2at^2}{t^2+2}, \frac{at^3}{t^2+2})$.Now the midpoint of $GP$ is the center $L$ which comes out to be,
$$\large x = \frac{at^4}{2(t^2+2)}$$
$$\large y = \frac{3at^3+4at}{2(t^2+2)}$$ 
Now, I am not able to eliminate $t$ from these equations. How do I proceed?  
Note: This problem is from SL Loney Co-ordinate geometry which was first printed a 100 years ago. So, there surely must be a shorter solution. 
 [/INST] We are given
\begin{gather}
\tag{1}\label{eq:1}
x = \frac{at^4}{2(t^2+2)}, \\
\tag{2}\label{eq:2}
y = \frac{3at^3+4at}{2(t^2+2)}.
\end{gather}
Combining \eqref{eq:1} and \eqref{eq:2},
$$
t\frac{y}{a} - \frac{x}{a} = \frac{2t^4 + 4t^2}{2(t^2+2)} = t^2,
$$
whence
\begin{equation}
\tag{3}\label{eq:3}
\left(t^2 + \frac{x}{a}\right)^2 = t^2\frac{y^2}{a^2}.
\end{equation}
Rewriting \eqref{eq:1},
$$
t^4 - 2\frac{x}{a}t^2 - 4\frac{x}{a} = 0,
$$
whence
\begin{equation}
\tag{4}\label{eq:4}
\left(t^2 - \frac{x}{a}\right)^2 = \frac{x^2}{a^2} + 4\frac{x}{a}.
\end{equation}
From \eqref{eq:3} and \eqref{eq:4},
$$
4t^2\frac{x}{a} = t^2\frac{y^2}{a^2} - \frac{x^2}{a^2} - 4\frac{x}{a},
$$
whence
\begin{equation}
\tag{5}\label{eq:5}
t^2\left(\frac{y^2}{a^2} - 4\frac{x}{a}\right) = \frac{x^2}{a^2} + 4\frac{x}{a}.
\end{equation}
With hindsight, this deduction could have been shortened, but I've preferred to keep it in a sequence that occurred fairly naturally. Anyway, we can now substitute \eqref{eq:5} in \eqref{eq:4}. How to do so, while making the least mess, is debatable. I decided it was about time to get rid of the denominators, thus
$$
(at^2 - x)^2 = x(x + 4a) = t^2(y^2 - 4ax),
$$
leading to this manageable mess,
\begin{align*}
4ax(y^2 - 4ax)^2 & = a^2[t^2(y^2 - 4ax)]^2 - 2ax(y^2 - 4ax)[t^2(y^2 - 4ax)] \\
& = a^2[x(x + 4a)]^2 - 2ax(y^2 - 4ax)[x(x + 4a)],
\end{align*}
whence
$$
\boxed{4(y^2 - 4ax)^2 = x(x + 4a)[a(x + 4a) - 2(y^2 - 4ax)]}
$$
Gathering all terms containing $y$ on the left hand side,
$$
4y^2(y^2 - 8ax) + 2x(x + 4a)y^2 = ax(x + 4a)[x + 4a + 8x] - 4(4ax)^2.
$$
Simplifying,
\begin{align*}
2y^2(2y^2 - 16ax + x^2 + 4ax) & = ax[(x + 4a)(9x + 4a) - 64ax] \\
& = ax(9x^2 - 24ax + 16a^2),
\end{align*}
and finally,
$$
\boxed{2y^2(2y^2 + x^2 - 12ax) = ax(3x - 4a)^2}
$$
</s>