Today has been a very busy day—I think I did every errand known to man—and I haven’t fit in any study time yet. I give my readers my solemn word to at least read some of The Joy of X before bed, however.
Today I am proving the proposition below for the case $a=2p$. The case $a=0$ was dealt with yesterday, and the cases $0<a<2p$ and $a>2p$ will be addressed tomorrow or later in the week.
Proposition: Consider the parabola $x^2=4py$, where $0<p$. Let $a$ be a real number. Let $A$ be the vertical line $x=a$. Let $B$ be the line through $F=(0, p)$, the focus of the parabola, and $R=(a,\frac{a^2}{4p})$, the point where $A$ intersects the parabola. Let $C$ be the line tangent to the parabola at $R$. Then, for all $a$, the acute or right angle formed by $A$ and $C$ is equal to the acute or right angle formed by $B$ and $C$.
First, note that the equation of the parabola can be written as $y=\frac{1}{4p}x^2$.
Now suppose $a=2p$.
Then, $R$ is the point $(2p, p)$, and $B$ is the horizontal line $y=p$.
Using calculus, $y’=(\frac{1}{4p})2x=\frac{1}{2p}x$. Thus, the slope of $C$, the line tangent to the parabola at $R=(2p, p)$, is $(\frac{1}{2p})(2p)=1$.
It follows that $C$ intersects $A$, $B$, and the x-axis as show in the diagram below.
In this diagram, $\alpha$ is the angle formed by $A$ and $C$ and $\beta$ is the angle formed by $B$ and $C$.
Note that $\alpha$ is equal to the angle that is vertical to $\alpha$.
Since $B$ is horizontal, it is parallel to the x-axis. Thus, $\beta$ is equal to the alternate interior angle to $\beta$ that is formed by $C$ and the x-axis.
Because $A$ is perpendicular to the x-axis, the triangle formed by $A$, $C$, and the x-axis is a right triangle. Hence the ratio of the legs must be equal to the slope of $C$, which is 1. It follows that the legs are equal.
Thus, the angle vertical to $\alpha$ and the alternate interior angle to $\beta$ are also equal, because they subtend the equal sides of an isosceles triangle.
Therefore, $\alpha=\beta$ and the proposition holds when $a=2p$.
(Fun!)
go-go