A special property when a straight line and a parabola intersect

Question

Consider the following graph:

The blue line indicates the maximum height from the line to the parabola. Let $P\equiv(x_0,0), Q\equiv(x_m,0)$ and $R\equiv(x_1,0)$. By looking at the coordinates we can find that $PQ=QR$. How can we prove it for the general situation?

(For further:
Parabola $\to y=\sqrt 3 x - \frac{x^2}{20}$
Line $\to y=\frac{x}{\sqrt{3}}$

I suppose these equations are not necessary. This is true for any parabola and line as above.)

^{(I asked this to get help on a question on PSE and my own answer for that)}

Answer 1

I do not see it so much as an analytic problem. It is a fundamental geometric property of parabolas. Your equation defines a parabola having a vertical axis. Let $PJ$ be the chord, and let $K$ be the point on the segment furthest from that chord. Construct line $t$, through $K$ and parallel to $PJ$. Let points $Q$ and $R$ lie on the $x$-axis, such that $KQ$ and $JR$ are both vertical, and let $KQ$ meet $PJ$ at point $L$.

Line $t$ can meet the parabola at only the one point. (Otherwise $K$ would not be the furthest point from the chord.) That makes $t$ tangent to the parabola. Line $KQ$ is vertical, making it a diameter of the parabola. Any chord parallel to $t$ is bisected by this diameter, so $L$ is the midpoint of $PJ$. This bisection property was covered by Archimedes in Quadrature of the Parabola (Proposition 1), and by Apollonius in Conica (I,46).

Line $KLQ$ is parallel to $JR$ and bisects $PJ$, so it must also bisect $PR$. That makes $Q$ the midpoint of $PR$.

Answer 2

The maximum distance from parabola to the line is at the point where the tangent to parabola is parallel to the line. The line's slope is $1/\sqrt 3$ and the derivative of the function defining a parabola: $$y=\sqrt 3 x - x^2/20$$ is $$y'= \sqrt 3 - x/10$$ which equals $1/\sqrt 3$ at $$x_m=10(\sqrt 3 - \tfrac 1{\sqrt 3})$$

The intersection of line and parabola is at $x$ where $$\sqrt 3 x - x^2/20 = \frac x{\sqrt3}$$ which yields $$x_0=0$$ or $$x_1=20(\sqrt3 - \tfrac 1{\sqrt 3})$$ hence $$x_m = \frac 12 x_1.$$

Answer 3

Since it is not a good thing to prove for a numerical example, I am here trying to prove it for a general situation, being inspired by the other answers.

Let the equations of parabola and straight line be $$y=ax^2+bx+c \tag{1}$$ and $$y=px+q \tag{2}$$ respectively.

Quoting CiaPan's answer,

The maximum distance from parabola to the line is at the point where the tangent to parabola is parallel to the line.

The derivative of the function of parabola is, $$y'=2ax+b$$

then, $$p=2ax_m+b$$ $$x_m =\frac{(p-b)}{2a}\tag{A}$$

From (1) and (2) we get, $$ax^2+(b-p)x+(c-q)=0$$ Roots of this equation give $x$-coordinates of the points of intersection of parabola and chord. Assuming this has real roots let's find the sum of two roots: $$\begin{align}x_0 + x_1 &= \frac{-(b-p)}{a}\\ &=\frac{(p-b)}{a}\tag{B}\end{align}$$

What we have to prove is $$x_m=\frac{x_0 + x_1 }{2}$$ Hence proved from (A) and (B).