Jul 292012
 

Many people ask me why I ride my motorcycle long distances in the summer. This summer I traveled from Reno, NV to St. Louis, MO. It was around 4000 miles, round trip, and brutally hot for a couple of states worth of riding.

But, that traveling allows me one single thing I rarely get. Time away from all distractions. It worked. I thought long and hard about the problem I talked about last post; A visual representation to imaginary solutions of quadratics. Somewhere in Wyoming I had the idea on how to prove it. By the time I hit Utah, I had the solution worked out in my head, and I needed to jot some notes. It honestly took me several hours to type up the solution, and without further ado, here it is.

graph5

The Goal:

To prove that in a general case, the circle that is created by reflecting a parabola with imaginary roots (the orange one) about its vertex (the black one) will have as its radius the value of the imaginary roots of the original.

We will begin with clip_image002[12] as our initial equation, with one requirement that the discriminant is negative;  clip_image004[10]. This will ensure that our initial quadratic equation has imaginary roots and the parabola exists above the x axis as shown.

 

Now, we need to reflect this equation around the vertex, but just adding a negative sign in front of the “a” will not do it. If we add that sign in and make it “-a” it will reflect around the x-axis, not the vertex. Therefore, we are going to need to complete the square, get the original equation in vertex form and then add the minus sign to reflect.

 

Given equation                                                                       clip_image002[13]

 

First, divide all terms by “a” and set the y = 0                              clip_image006 

This gives us a first coefficient of 1, which makes

Completing the square possible. Next, we will complete

the square by using clip_image008 and its square.                        clip_image010

 

now that the perfect square trinomial has been constructed       clip_image012

we can factor the trinomial into vertex form.

 

The center of the circle above can be clearly seen in this form, and is: clip_image014 We will need this later.

 

Now we need to solve the reflected parabola for x.               clip_image016

 

Add & Subtract the constant terms from both sides to get:          clip_image018

 

Move the negative sign from the right to the left side:                 clip_image020

 

Take the square root of both sides:                                       clip_image022

 

Finally subtract the constant term from both sides:                  clip_image024


Notice that we have essentially derived a version of the quadratic formula. It doesn’t look exactly like the standard version we all memorize, but it is the same, with one important difference. There is a sign change to the terms inside the radical sign! That will be very important.

 

This formula gives us where the reflected parabola crosses the x-axis, so we now have 2 points on the circle, the plus and minus, and the center of the circle.

 

The final step of the proof is to show that the radius of the circle, or to put it in another way, the distance from the center of the circle to one of the roots of the reflected parabola, is identical to the imaginary part of the solution / roots of the non-inverted parabola. So, onward to the distance formula.

 

We need to find the distance from clip_image014[1] to clip_image026.

 

Distance formula:                                            clip_image028

Insert the point values for x and y       clip_image030


Using just 1 of the 2 values for the + or -.

 

Simplify the subtractions:                                               clip_image032

 

Finally, square the inside term leaving the following:             clip_image034

 

This leaves us with a pseudo-determinant of:                     clip_image036

 

 

However, in setting up the problem initially, we stipulated that the determinant clip_image038 would be negative. If that is true, then the value of inside the radical sign in our last step must be positive!

 

[And yes, I am cheating. I am leaving it to the reader to show that the way it is written above in the last step as the “pseudo-determinant” and the regular determinant are essentially equivalent.]

 

Not only that, but the value of clip_image040 which is from our inverted quadratic, is the same value but opposite sign of the more familiar clip_image042 from the quadratic equation.  If clip_image038[1] is negative, our inverted quadratic will be positive with the same value (oh, and it works in reverse too!)

 

There, I now proved that the reflecting a parabola with imaginary roots around its vertex will allow you to calculate the imaginary part of the complex answer as the radius of a circle created by the reflection.

 

QED.

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