Nothing in my brain is really working today. I read two chapters of The Joy of X, but focusing enough to follow them was very difficult. I’m definitely not up to posting more about parabolas at the moment. Sigh.
Month: March 2024
Parabola-a-Go-Go, Part 2
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!)
Parabola-a-Go-Go, Part 1
Today I will begin sharing the results of my investigation into reflections in parabolas. For a refresher on the problem, see my post Parabolic Reflections.
First of all, I have chosen to investigate only upward-facing parabolas with their verticies at the origin. If the conjecture is true for those parabolas, then it is true for parabolas in general, because the lines involved in the problem will form the same angles if translated, rotated, or reflected.
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 $C$.
The relationships can bee seen in the figure below.
I found it necessary to split my proof into cases, depending on the value of $a$. The figure above represents the case $a=2p$. In that case, the slope of $B$ is zero. The other relevant cases are $a>2p$, where the slope of $B$ is positive, $0<a<2p$, where the slope of $B$ is negative, and $a=0$, where the slope of $B$ is undefined. The last of these cases is pictured below. As you can see, the proposition is trivially true when $a=0$ because $A$ and $B$ are the same line.
I will not discuss the cases where $a<0$. Because a parabola is symmetrical, it is sufficient to prove that the proposition holds for one half of it. (I did, in fact, consider those cases. I initially did the proof for the left side of the parabola instead of the right side. The math for the right side is slightly simpler, though, so that is what I will be presenting here.)
Tune in tomorrow for the proof for the case $a=2p$. It’s quite simple if you know that the slope of the line tangent to the parabola at $x$ is $\frac{x}{2p}$. Readers might enjoy seeing if they can work it out for themselves.
Reticulating Splines
Today I finished my investigation of reflections in parabolas. I also made some of the diagrams I will need when writing up my results. Yet I was tired and listless all day and, when it came to actually composing a post, I hit a wall. I hope for better conditions tomorrow.
Working, Please Wait
I spent both some wakeful time last night and all of my study time today working on my parabola problem. It’s turning out to be an ideal project for this point in my review, as I’m using all the subjects I’ve been learning—algebra, analytic geometry, conics, and trigonometry—plus a bit of remembered calculus.
One thing I did today was some research on inverse trigonometric functions. In the process, I discovered a resource called LibreTexts, which offers a largish selection of free math textbooks. I found it useful to look at a couple of different treatments of the same topic.