# Proving that the focus of a parabola lies on the circumcentre of a triangle

I recently came across the fact that if a parabola touches the three sides of a triangle then the focus of such a parabola lies on the circumcircle of the above triangle.

I tried to prove it but without much information I couldnot get where to start with .Does the above property of the parabola applies to other conic sections as well? Any help is appreciated.Thanks.

• I don't have the proof, but I don't see how a parabola could touch the three sides of a triangle. My feeling is that if a parabola touches two sides of a triangle, it intersect the other at two points. – ajotatxe Mar 22 '17 at 16:50
• Perhaps the intended interpretation is that the lines on which the sides of the triangle lie are all tangent to the parabola. For example, with the parabola $y=x^2$, the triangle with vertices $(1,1)$, $(-1,1)$, and $(0,-1)$ has this property (and I believe its circumcenter is indeed the focus $(0,1)$ of the parabola). – Greg Martin Mar 22 '17 at 17:21
• @GregMartin This parabola is not tangent to the line $y=1$ which is one of the sides. – Wojowu Mar 25 '17 at 14:05
• Are you sure about this statement? One can show that the focus lies on the circumcircle, but will never coincide with circumcenter. – Wojowu Mar 25 '17 at 14:05
• @Wojowu sorry you are right. I will edit my question. How can one show that focus lies on circumcircle then? Thanks. – Navin Mar 25 '17 at 14:27

Short Proof:

You need two lemmas:

1. The foot of the perpendicular from the focus to a tangent of the parabola lies on the tangent at the vertex.That means the feet of the perpendicular to the three sides of the triangle formed by the tangents lie on a straight line, called the Simson Line which leads us to use

2. Simson-Wallace Theorem: The Feet of the perpendiculars from a point to the sides of a triangle are collinear iff the point lies on the circumcircle.

It now follows that focus lies on the circumcircle

• Thanks for your answer.This implies that the fact applies to ellipse and hyperbola as well. – Navin Mar 28 '17 at 7:25
• In those cases the foot of the perpendicular will lie on the auxiliary circle instead of on a st line. So it will not apply. – Hari Shankar Mar 28 '17 at 10:04
• No I mean that it is a common property of ellipse,hyperbola and parabola that the foot of the perpendicular from focus to any tangent to the conic lies on the tangent at the corresponding vertex.This implies that the fact works well for ellipse and hyperbola as well. – Navin Mar 29 '17 at 17:07
• Where can I get such lemma s or theorems? How to prove this Simson Wallace theorem? – Alex Jul 30 '20 at 20:18

Another way: Let the three lines be tangent at $t_1, t_2, t_3$ on the parabola. Then the points of intersection of these tangents, which form the vertices of the triangle are $A(at_1t_2, a(t_1+t_2))$ etc.

You can then show that these points along with $(a,0)$ form a cyclic quadrilateral, by using slopes to show that opp angles are supplementary.

)The following works, but is not very inspiring:

Take the parabola as $y=x^2$ with focus $F$ at $\bigl(0,{1\over4}\bigr)$. The tangent $t_a$ at $(a,a^2)$ has equation $y=2ax-a^2$. Intersecting $t_a$ with $t_b$ gives the vertex $C=\left({a+b\over2},ab\right)$ of the triangle. By symmetry we then have $A=\left({b+c\over2},bc\right)$ and $B=\left({c+a\over2},ca\right)$. Further computation leads to the circumcenter $$M=\left({1\over4}(a+b+c-4abc), \ {1\over8}(1+4ab+4bc+4ca)\right)\ ,$$ and the circumradius $R$ satisfies $$R^2={1\over64}(1+4a^2)(1+4b^2)(1+4c^2)\ .$$ It is now easy to verify that $F$ is lying on the circumcircle of $\triangle(ABC)$.