Mathosaurus

In response to a question on Khan Academy, here is one way to show that the formula

is an ellipse. The idea is this: rather than trying to understand (1), I will define an ellipse from scratch and provide a construction, convert them to algebra, and test equation (1) for myself.

I. SET UP THE PROBLEM

ASSUMPTIONS:
• I accept the the theorems from the Khan Academy Geometry playlist, and the rules of arithmetic and Algebra.
• Given two arbitrary fixed points A and B; and constant length s > AB;
• Place point P so that the sum of lengths AP + BP = s.

DEFINITION: An ellipse is the figure traced by point P.
Or, An ellipse is the locus of all points P such that AP+ BP = s,
where A, B are fixed points, s an arbitrary constant > AB.

PROPOSITION: To draw an ellipse.
Or, to find the locus of P.

CONSTRUCTION:
I will use a piece of paper, a pencil, two tacks, and a length of string.
Cut the string to any length s, and attach one end to each tack. Fix tacks A and B to a sheet of paper so they don't move, and leaving at least a little slack in the string. Name the pencil P and talk sweet to it. Pull the string taut with the tip. Draw, keeping the pencil on the page, and the string tight.

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Fig. 1

I say, the figure drawn is an ellipse.

For the sum AP + BP is the string length s, and points A, B are fixed.

"I say..." is a bit silly because there's nothing to prove. I'm looking for mistakes. Next I say, "I don't believe you," and check.

II. CONVERT THE PROBLEM TO ALGEBRA

DESCRIBE THE FIGURE
• On triangle ΔABP, let AB = c, AP = d, BP = e.
• Let the line through A,B be the x-axis, and the midpoint of AB be the coordinate origin O = (0,0).
The coordinates of the tacks are then A = ( c/2, 0), B = (-c/2, 0)
• Drop a perpendicular from P to line AB, and
• Let the coordinates of P be (x, y).
I should also pick up the major and minor axes.
• Major Axis: The figure is widest along line AB. The width is s.
(Proof: The sum d + e is maximum when triangle ΔAPB vanishes to a line http://www.khanacademy.org/math/geometry/triangle-properties/e/e/triangle_inequality_theorem )
• Minor Axis: The figure is narrowest on the perpendicular bisector of AB, when H=O, and d = e.
(Proof: Use the Law of Cosines.)

Let a and b be the radii on the major and minor axes.

From ΔAOP = ΔBOP,

Fig.2 (try clicking)

Now I want to give the value of y for any x.

BUILD AN EQUATION IN X AND Y:
I will pick x by moving H from side to side. With O as the origin, x is negative to the left of O. Similarly, positive distances (arrows) point right, negative ones point left. Let u = HA, v =BH (Fig. 3) be the bases of two right triangles ΔAHP, ΔBHP (Fig. 3). Minding the sign of x, I have

Addition and subtraction of segments can be tricky; let me know if this is confusing.
Also by the Pythagorean Theorem,

I have u, v in terms of c, s, and x. Let me do the same for d. In (3), subtract the second equation from the first, eliminating y:

Replace e, with s−d, u, v with the values in (2):

Expanding the parentheses, cancelling a bunch of squares, and solving for d, I get:

Plug this value of d back into the first equation of (3). It now reads,

or,

Fig.3

This is an expression in x and y, with two arbitrary constants. I can now ask my question.

III. CHECK

PROPOSITION: Equation (1) is an ellipse. That is,

For any two given values of a and b, there are two fixed points A=(c/2, 0) and B=(− c/2, 0), so that for every point P(x, y) that is a solution of (1), the sum of lengths AP+ BP is a constant s: d + e = s.

I also assume a and b are not zero since division by zero is undefined.

I know that equation (5) is an ellipse. Let me be grumpy about that. No amount of math can stop me from building an ellipse and turning it into (5).
Proof: there are such things as string and tacks, and I know where to get them.

[Soapbox]

That was fun. It's also the most important step. There is a part in The Little Prince when he makes a drawing:

Nice hat. But it's not a hat. It's a boa constrictor eating an elephant.

....see?
The first drawing... that's Algebra. The shape is the same no matter what's inside, or how it's made. Algebra is fast, flexible and powerful.
But usually I am stuck because I don't understand. This not-understanding is unlike anything else. Substitute the value of x from the last step and then prog the blat poodlemath msapdorif scrbab math zittle. bork.
Wait. what?

When this happens I need more information. What am I doing and how does it work? One way to figure out what's happening is to make a more detailed picture.

Draw an elephant. You can always color over it with a brown marker later.

Start with Geometry. You can always make it abstract later. (convert it to Algebra or Calculus).

[End Soapbox]

I have the following
TEST:
If x²/a² + y²/b² = 1 is an ellipse, then I must be able to turn (5) into (1) using the rules of ordinary Algebra.

Begin. Square both sides to eliminate the radical, and multiply out the squares.