A Mathematical Beauty of Hyperbolas

This morning in my Algebra 2 Honors class, my teacher decided to show my class and I some extra tid-bits. These tid-bits were perhaps one of the most beautiful mathematical concepts I ever saw. We were learning about analytic geometry and hyperbolas at that time. For those of you who don’t know what a hyperbola is, check this graph below:

To sum it up in general terms, a hyperbola is pretty much any graph that looks like the one above- the red curves only-(vertical or horizontal). However, there is one property of the hyperbolas that I will focus on for this post: notice that besides red curves, there are black lines also. These black lines are asymptotes, in which the graph of the hyperbola- the red curves- approach them but never ever touch them.

Visually, yes, we understand this property. However, what my math teacher showed me this morning was how to explain this in mathematical terms, specifically in limits. Isn’t that almost magical? The fact that you can explain a visual quality in numbers and equations?

Anyway, let us first take take the equation of this hyperbola in the graph above.

 x²/16 – y²/9 = 1

Notice that in order to illustrate this relationship in which the graph approaches but never touches the asymptotes, y has to be near positive/negative infinity. So let us get the y-coordinates of this equation. To do so, simply solve y in terms of x using algebra. This will result in-

                                         y=±¾ √(x²-16)

   [which equals by factoring out an x² in the radical..] y=±¾ √[x²(1 –  16/x²)]

   [which equals by taking the x^2 out of the radical..] y=±¾x √(1 –  16/x²)

 Notice again that for this relationship to occur not only does y have to near pos/neg infinity, so does x. So suppose x is approaching negative or positive infinity. Or in math, x→±∞. Let us using this value plug x into the equation that we have now. This will cause 16/x² to be approximately equal to 0, because 16, a small number, out of infinity, a super large number, is a really small fraction that has to be very very close to zero. In math terms, 16/x²→0.

Now, let us get this value for 16/x^2 and plug it back into the equation. This will cause the terms under the radical to go like this:

Since 16/x²→0, (1 –  16/x²)→ (1-0) → 1

Substitute this (1 –  16/x²)→ 1 into the equation y=±¾x √(1 –  16/x²) and get-

y→±¾x √(1 –  16/x²)→ ±¾x √(1)→ ±¾x

given x→±∞ ; y→±¾x

Tada! Let’s look at this limit equation we got. First of all, refer back to the graph. Notice that the ±¾x  is in fact the equations of the asymptote lines. So what this limit equation is basically saying is that as x approaches negative/positive infinity, y or the function will approach  ±¾x aka the asymptotes. In other words, we have just mathematically illustrated the asymptotic relationship in which the hyperbola will near but never touch the asymptotes. Not just that, we in a sense derived it from a hyperbolic equation; thus we proved that in any hyperbola, the graph will always have this relationship. Isn’t that beautiful?

Perhaps the way I’m showing it to you on the computer is not as beautiful as it really is. (It’s actually pretty hard to write these equations on a computer. I couldn’t even use Words for this.) I would agree that showing this on paper and actually writing this out makes this derivation seem so much more beautiful.

But what makes this beautiful, computer or not? I dunno really, but I think it is the fact that it is so simple and so elegant. Just a small number of steps and you can explain a visual property in mathematical terms. I don’t know about you, but whenever I read through this process, there is a thrill down my spine. It is just so mathematically beautiful.

Advertisements

One thought on “A Mathematical Beauty of Hyperbolas

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s