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Parabola-a-Go-Go, Part 2

Olly / / Math Posts

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.

A Equals 2p Annotated 1

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

Olly / / Math Posts

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.

A Equals 2p

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.

A Equals 0

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.

Working, Please Wait

Olly / / Math Posts

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.

Dancing Graphs

Olly / / Math Posts

Today I was tired again and not feeling mathematically creative, so I worked on exercises in the review of trigonometry rather than doing more work on the parabola problem. A few of the exercises involved graphing trigonometric functions. I’ve always found something enchanting in the graphs of sine and cosine, especially when they dance together as shown below.

Simultaneous Sine And Cosine

I’ve also always wondered if there is a name for the leaf-like shape formed between them

Shape Between

(And while we’re on the subject, check out this lovely animation I found of sine and cosine graphed side by side.)

Sine And Cosine

Source

Parabolic Reflections

Olly / / Math Posts

Today I spent all of my study time investigating the parabola problem from last week. There are still details to be worked out, but I’m fairly sure the approach I’m using will allow me to prove that the angles I conjectured might be equal actually are. I thought today I’d give a better explanation of what my conjecture was.

A well known principle in physics, called the Law of Reflection, is that when light reflects from a flat surface, the angle of incidence (that is, the angle at which the light hits the surface) is equal to the angle of reflection (that is, the angle at which the reflected light leaves the surface). This is shown in the diagram below.

Angles Of Incidence And Reflection 1

A well known property of parabolas is that, if light is emitted from the focus of the parabola ($F$ in the diagram below), then it will reflect from the surface of the parabola in a direction parallel to the parabola’s orientation (vertical, in the diagram). This is why parabolic mirrors are used in car headlights to reflect all light outward from a bulb at the focus.1

Reflecting Parabola

My question is basically how this property of parabolas relates to the Law of Reflection. It is not obvious how one would measure the angle at which light hits or reflects from a curved surface. What would that mean? One possibility is to measure the angle between the path of the light and the line tangent to the curve at the point of reflection. (In the diagram, the point of reflection is called $P$ and the blue line is tangent to the parabola at $P$.2)

My conjecture is that the a generalization of the Law of Reflection holds true for light emitted from the focus of the parabola and reflected at $P$, if one interprets the angles of incidence and reflection as the angles formed with the line tangent to the parabola at $P$. That is, I conjectured that the angle between the blue tangent line in the diagram and the vertical line through point $P$ is the same as the angle between the blue tangent line and the segment $\overline{PF}$.

  1. Moving in the other direction, if light hits the surface of the parabola from a direction parallel to the parabola’s orientation, it will reflect along a line that passes through the focus of the parabola. This is why parabolic mirrors are used in power generation to concentrate all incoming light on a collector a the focus. ↩︎
  2. The line tangent to a parabola at a point intersects the parabola at that point and no other, and its slope is sometimes described as the slope of the parabola at that point. ↩︎

Rubber Ball

Olly / / Math Posts

I was still very tired for most of today, but I rallied in the late afternoon and managed to do 90 minutes of trig review and 30 of investigation into the parabola problem. I think I’m making progress, though I haven’t started trying to work out all the algebra and trig yet.

I will probably need some inverse trigonometric identities. I was not certain such things existed, but apparently they do. Most of the exercises I did today concerned the regular kind: either proving them or using them to turn one expression into another. I remember doing a lot of the latter kind of problem in my tenth-grade precalculus class. I always thought they were kind of fun.

A Slow Day

Olly / / Updates

Today I felt almost normal again. I also felt terribly weary. I nearly didn’t do any math at all, but pushed myself to read a few chapters of The Joy of X after dinner. It would be a shame to have kept up with my daily study through all the physical and mental turbulence brought on by the change in my medication, only to falter when things started to smooth out again.

The Joy of X is all right. It deals shallowly with a wide range of topics, and I often find myself wishing for more substance. It is useful on days when I am not up to other types of review, however, and it has also pointed out some interesting avenues of inquiry.

Unfocus Meets Focus

Olly / / Math Posts

Today was rocky, though I am feeling a little better now. My math activity for the day was more reading in The Joy of X. The section I read included discussion of an imagined elliptical pool table on which a ball shot from one focus will always fall into a hole at the other focus. Since I know that pool balls (at least ideal ones) bounce off the cushion at the same angle at which they hit it, that had me wondering whether line segments like those shown below in red must always form equal angles with the line tangent to the ellipse at the point where they intersect. (The tangent line is shown in blue.) This seems likely, but I’m not sure how I would go about investigating it.1

Ellipse With Tangent

Naturally, I then wondered a similar thing about parabolas, so useful for reflecting and focusing light, which behaves in the same way as ideal pool balls.

Parabola With Tangent Two

(Sorry the explanation of these questions is a bit sketchy. I know it may be difficult for those without a lot of math background to understand, when this is a topic that should be accessible to them. This is the best I can do at the moment, though.)

  1. To do. ↩︎