Two equal parabolas with foci at S and S' touch each other at point P, such that PS'=PS. If the parabola with focus S is fixed, find the locus of S'

Hint given: Let the common tangent be L. PS and PS' are equally inclined to L which implies that P,S,S' are collinear or SS' is perpendicular to L. Thus the locus will be a parabola or directrix of a fixed parabola

My attempt: enter image description here

I can't figure out how to prove that PS and PS' are equally inclined to L. I've assumed the fixed parabola to be a standard one, with vertex at the origin. If we join S and S' and construct a perpendicular from P to SS', we can prove the two triangles are congruent. Thus S,S',P have to be collinear, or SS' is perpendicular to L.

Case 1:S,S',P are collinear

enter image description here

Assume P to be $(at^2,2at)$ and S to be (h,k)

Then, $(h-a)^2+k^2=2a(t+1)^2$, where t is the parameter. This looks like it will give a parabola upon plotting the locus, but I can't think of a proof.

Case 2: SS' is perpendicular to L

Then SS' will be the normal to the parabola, and the mirror image of the focus of the parabola about the tangent lies on the directrix. So the locus o S' in this case will be a straight line.

  • $\begingroup$ What is a parabola for you? $\endgroup$ Sep 19, 2021 at 7:28
  • 1
    $\begingroup$ @FedericoFallucca a conic section with an eccentricity of 1 or the locus of a point which has the same distance from a fixed point (focus) and a fixed line (directrix) $\endgroup$
    – Tatai
    Sep 19, 2021 at 7:35
  • $\begingroup$ See also this answer: math.stackexchange.com/questions/2562290/… $\endgroup$ Sep 19, 2021 at 10:13
  • 1
    $\begingroup$ @ Sunaina: Is the second parabola rolling on the first? What is changing? Is the topic is about " Roulette (curve)"? Wiki gives the locus as Cissoid in: en.wikipedia.org/wiki/Roulette_(curve) $\endgroup$
    – Narasimham
    Sep 19, 2021 at 17:14
  • $\begingroup$ @Narasimham No, it is not a cissoid but a parabola as said in the question and well proven by Math Lover. A cissoid is generated for example in the case of a parabola rolling on a straight line (mathcurve.com/courbes2d.gb/cissoiddroite/cissoiddroite.shtml) $\endgroup$
    – Jean Marie
    Sep 19, 2021 at 20:30

1 Answer 1


Let's take parabola $y^2 = 4 ax$ as the fixed parabola. The point you need to note is that both parabola are equal. In other words, the other parabola is $y^2 = 4 ax$ but rotated and / or translated.

On your question as to why the angles are same while you can show using algebra it is not really necessary. Here is some insight that may help. There are two ways to get the common tangent -

First - you fix point $P$ on the original parabola $y^2 = 4 ax$ and then rotate it around point $P$ till the tangents at $P$ align. The new parabola will be opening up in the negative x-direction.
Second - we take $P'$ on the original parabola such that it is mirror image of point $P$ about x-axis. We now shift the parabola such that $P'$ comes to point $P$ and then rotate the parabola counter-clockwise till tangents align and the new parabola also opens towards positive x-direction.

enter image description here

As mentioned above, there will be two equal parabola at a given point sharing the common tangent - one that opens towards positive x direction and the other towards negative x direction. $S, P$ and $S'$ will be collinear for the parabola that opens towards negative x-direction and $S S'$ will be perpendicular to the tangent line for the parabola that opens towards positive x-direction.

Case $a)$ - $S, P$ and $S'$ are collinear

As $P (a t^2, 2 a t)$ is the midpoint of segment $SS'$,

$at^2 = \frac{a + x_0}{2}, 2 at = \frac{y_0}{2}$

So coordinates of $S'$ is $(2at^2 -a, 4 a t)$ and the locus can be written as,

$x = 2 a \cdot \left(\frac{y}{4a}\right)^2 - a \implies y^2 = 8 a (x + a) $

For case $b)$, you can geometrically show it or find the intersection of the tangent line at $P(at^2, 2at)$ and the perpendicular line through $S$. As the intersection is the midpoint, you can find coordinates of $S'$ which comes to $(-a, 2 a t)$ and that is directrix of the fixed parabola.

  • $\begingroup$ "As the relative position of the point on the parabola is the same, it must make equal angle to the tangent line" I couldn't understand this line, the lengths will be equal, but how is this related to the angle of the parabola? $\endgroup$
    – Tatai
    Sep 19, 2021 at 7:52
  • 1
    $\begingroup$ @Sunaina parabolas are equal. In other words, take the parabola $y^2 = 4 ax$ and draw a tangent at a point. Now if we rotate the parabola and shift, will the angle between the tangent line and the $SP$ change? $\endgroup$
    – Math Lover
    Sep 19, 2021 at 7:58
  • 1
    $\begingroup$ @Sunaina I added some details. See if it helps. $\endgroup$
    – Math Lover
    Sep 19, 2021 at 8:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .