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I was reading about the properties of parabola, amongst which one of the property was that parabola has no centre.

I tried to prove it by considering four parametric points on the parabola i.e. $P_1(a(t_1)^2,2at_1), \,P_2(a(t_2)^2,2at_2), \\P_3(a(t_3)^2,2at_3), \,P_4(a(t_4)^2,2at_4)$

Further I equated the coordinates of midpoint of $P_1P_2$ and $P_3P_4$, after doing this I got that either $P1=P3$ and $P_2=P_4$ or $P_1=P_4$ and $P_2=P_3$, i.e. the two chords are coincident .

So from the above observation can I conclude that for a parabola a point which lies inside the parabola cannot be the midpoint of more than one chord?

SarGe
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Sameer Nilkhan
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    Assuming all your working is correct, then yes, this is a valid interpretation of the given conclusion. The point is that ${P_1, P_2} = {P_3, P_4}$, as unordered sets, showing that there is a unique unordered pair of points whose midpoint is the given point. – user803264 Jun 29 '20 at 03:11

2 Answers2

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Let $(x,y)$ be a point satisfying $y > x^2$. Then the system $$\begin{align} x &= \frac{x_1 + x_2}{2}, \\ y &= \frac{x_1^2 + x_2^2}{2} \end{align}$$ has, up to permutation of $x_1, x_2$, the unique solution $$x_1 = x \pm \sqrt{y-x^2}, \quad x_2 = x \mp \sqrt{y-x^2}.$$ It follows that for each "interior" point of the parabola $y = x^2$, there is precisely one pair of points $(x_1, x_1^2)$, $(x_2, x_2^2)$ on the parabola whose midpoint is $(x,y)$. Since all nondegenerate parabolas are similar, the result follows.

heropup
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A well-known property of parabolas states that

the midpoints of parallel chords all lie on a line parallel to the axis, and the tangent at the intersection point between that line and the parabola is parallel to the chords.

This implies that two different chords cannot have the same midpoint.

Intelligenti pauca
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