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
Locus of center

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.

  • $\begingroup$ The answer given at the back of the book is $2y^2(2y^2+x^2-12ax) = ax(3x-4a)^2$. $\endgroup$ – Shubhraneel Pal Nov 16 '18 at 14:59
  • 1
    $\begingroup$ You can ask Wolfram Alpha to eliminate $t$ from your equations via the command Eliminate[ x == a t^4/(2 t^2 + 4) && y == (3 a t^3 + 4 a t)/(2 t^2 + 4), t]. The result you get is equivalent to the one you cite from the back of the book, so evidently it is possible to eliminate $t$ from the above. (It does seem laborious, however, and it may be that there's a better approach.) $\endgroup$ – Semiclassical Nov 16 '18 at 15:28

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, $$ \boxed{2y^2(2y^2 + x^2 - 12ax) = ax(3x - 4a)^2} $$

  • $\begingroup$ Taking sum and difference of squares, completing squares etc. are common methods of elimination, I think this is exactly what the book expected. $\endgroup$ – Shubhraneel Pal Nov 17 '18 at 1:21

Let us find the slope of the tangent:




Distances $OL$ and $LP$ are equal:


The point $P$ lies on the parabola:


Line $LP$ is tangent to parabola:


Introduce the following substitutions:



Replace (4) and (5) into (2) and (3):



From (6) and (7) you get:



Rearrange (1):





Now replace (8) into (10) and the final result is the locus of points $L(x_L,y_L)$.

Hopefully I did not make any mistake along the way. Even if I did, I think that I have demonstrated the right way.

  • $\begingroup$ Yeah but the simplification of the expression is way too messy just as would have been with my method. Although, in this method the equation is easier arrived at than my method which needs a lot of indirect approach to eliminate $\endgroup$ – Shubhraneel Pal Nov 17 '18 at 1:19

Simplify slightly by substituting $x = au/2$ and $y= av/2$. Then your parameterization yields $$u(t^2+2)=t^4 \qquad v(t^2+2)=t(3t^2+4) \tag{1}$$ Squaring the second equation gives us a system in even powers of $t$, so define $s := t^2$, giving $$u(s+2) = s^2 \qquad v^2(s+2)^2=s(3s+4)^2 \tag{2}$$ We can use the $u$ equation to whittle-down powers of $s$ in the $v$ equation. It'll be tedious, but completely mechanical. At least we can look forward to the fact that the end result will be linear in $s$.

$$\begin{align} v^2(\;s^2+4s+4\;)&=s(\;9s^2+24s+16\;) \tag{3}\\ v^2(\;u(s+2)+4s+4\;)&=s(\;9\cdot u(s+2)+24s+16\;) \tag{4}\\ v^2(\;s(u+4)+2(u+2)\;)&=3s^2(3u+8)+2s(9u+8) \tag{5}\\ v^2(\;s(u+4)+2(u+2)\;)&=3(3u+8)\cdot u(s+2)+2s(9u+8) \tag{6}\\ s(\;v^2(u+4)-9u^2-42u-16\;) &= 6u(3u+8)-2v^2(u+2) \tag{7} \end{align}$$

From here, we ---and by "we", I mean "the reader"--- can solve $(7)$ for $s$ and substitute back into the $u$ equation in $(2)$. Once the dust settles, "we" will have the $u$-$v$ equivalent of the target relation:

$$v^2 (\;2v^2+u^2-24 u\;) = u (\;3u-8\;)^2 \tag{$\star$}$$


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