Food for thought

## Solution

Take any two points $A$ and $B$ on the parabola $y = x^2$.

Draw the line $OC$ through the origin, parallel to $AB$, cutting the parabola again at $C$.

Let $A$ have coordinates $(a,a^2)$, let $B$ have coordinates $(b,b^2)$ and let $C$ have coordinates $(c,c^2)$.

Prove that $a+b = c$.

We know that

$A=(a,a^2)$,

$B=(b,b^2)$,

$C=(c,c^2)$.

We can calculate the gradient of a straight line between two points $(x_1,y_1)$ and $(x_2,y_2)$ as $\frac{y_1-y_2}{x_1-x_2}$.

So the gradient of line $AB$ is equal to $\frac{a^2-b^2}{a-b}=a+b$.

Also, the gradient of line $OC$ is equal to $\frac{0-c^2}{0-c}=c$.

However, we know from the question that these two lines are parallel. Two lines are parallel if and only if they have the same gradient, which means that we must have $a+b=c$.

Imagine drawing another parallel line $DE$, where $D$ and $E$ are two other points on the parabola. Extend the ideas of the previous result to prove that the midpoints of each of the three parallel lines lie on a straight line.

Now, suppose that we have two points $D=(d,d^2)$ and $E=(e,e^2)$ on the parabola making another line parallel to $OC$.

The gradient of line $DE$ is equal to $\frac{d^2-e^2}{d-e}=d+e$. As $DE$ is parallel to $OC$, we also have $d+e=c$.

We can calculate the midpoint of a straight line between two points $(x_1,y_1)$ and $(x_2,y_2)$ as $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$.

So the midpoint of line $OC$ is $\left(\frac{0+c}{2},\frac{0+c^2}{2}\right) = \left(\frac{c}{2},\frac{c^2}{2}\right).$

The midpoint of line $AB$ is $\left(\frac{a+b}{2},\frac{a^2+b^2}{2}\right) = \left(\frac{c}{2},\frac{a^2+b^2}{2}\right).$

The midpoint of line $DE$ is $\left(\frac{d+e}{2},\frac{d^2+e^2}{2}\right) = \left(\frac{c}{2},\frac{d^2+e^2}{2}\right).$

All three of these points have an $x$-coordinate of $\frac{c}{2}$. This means that the line $x=\frac{c}{2}$ passes through the midpoints of the lines $OC$, $AB$ and $DE$.