Show that the baricenter of the $4$ intersection points of a parabola with a circle is on the axis of the parabola.

Let $p$ be a parabola, $c$ a circle and $p\cap c=\{P_1,P_2,P_3,P_4\} \Rightarrow B= \frac{P_1+P_2+P_3+P_4}{4} \in$ axis of $p$.

  • $\begingroup$ Example. $y=x^2;(x+\frac{9}{8})^2+(y-\frac{27}{8})^2=\frac{325}{32}$ 4 solutions: $(1,1), (2,4), (-\frac{1}{2},\frac{1}{4}), (-\frac{5}{2},\frac{25}{4})$ The baricenter of these 4 points is: $B = \frac{1}{4}( (1,1)+(2,4)+(-\frac{1}{2},\frac{1}{4})+(-\frac{5}{2},\frac{25}{4}))= (0, \frac{23}{4})$ so $B$ is in the line $x = 0$ which is the axis of the parabola. $\endgroup$
    – Sgernesto
    Feb 7 '13 at 20:18

Without loss of generality we can take the parabola as having equation $y=kx^2$.

Write down the equation $(x-a)^2+(y-b)^2=r^2$ of the circle. Substitute $kx^2$ for $y$.

You will get a degree $4$ polynomial in $x$ with no $x^3$ term.

But the sum of the roots of the polynomial is the negative of the coefficient of $x^3$, divided by the coefficient of $x^4$.

So the sum of the four (not necessarily real) roots is $0$. The $x$-coordinate of the barycenter of the points of intersection is, if there are four real roots, not necessarily distinct, the mean of these $4$ roots.

  • $\begingroup$ "Great minds think alike." $\endgroup$ Feb 7 '13 at 20:44
  • $\begingroup$ Well, I am not sure I did much thinking. Have assigned problems of this general kind in the past. $\endgroup$ Feb 7 '13 at 20:55
  • $\begingroup$ The motivation for this is a rule and compass question: given a parabola construct its focus and directrix. So now we know how to construct a point in the axis and using it we construct what is asked. $\endgroup$
    – Sgernesto
    Feb 7 '13 at 21:05

If we expand a fourth-degree polynomial function $x\mapsto(x-p)(x-q)(x-r)(x-s)$, we get $$ x^4 - (p+q+r+s)x^3 + \cdots. $$ The sum of the roots is minus the coefficient of $x^3$.

The $x$-coordinates of the intersection of the parabola $y=x^2$ with the circle $(x-h)^2+(y-k)^2 = r^2$ are the roots of $$ (x-h)^2 + (x^2-k)^2 = r^2. $$ That is a fourth-degree polynomial in $x$. The coefficient of $x^3$ is $0$. Therefore the sum of the roots is $0$.

  • 1
    $\begingroup$ I see someone up-voted this answer and someone else down-voted it. It's essentially the same as the independent answer posted by André Nicolas, which got five up-votes and an acceptance. Is there something objectionable about my answer? $\endgroup$ Feb 9 '13 at 0:08

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