Equation of the ellipse

Here is a simple derivation of the equation for an ellipse. It is taken from Morris Kline's book: Calculus, An intuitive and Physical Approach.

We place the two foci of an ellipse (F and F') at the coordinates (c,0) and (-c,0). Each point on the ellipse is defined by the property that the sum of the distances to F and to F' is constant, which we define as equal to 2a. Our old friend Pythagoras helps us find the distances in terms of x,y and c.

PF = √[y2 + (x-c)2] PF' = √[y2 + (x+c)2] 2a = PF + PF' PF = 2a - PF' √[y2 + (x-c)2] = 2a - √[y2 + (x+c)2]

Square both sides (and expand):

y2 + x2 - 2xc + c2 = 4a2 - 4a √[y2 + (x+c)2] + y2 + x2 + 2xc + c2

Cancel terms (x2, y2, and c2) and rearrange to isolate the remaining square root:

- 2xc = 4a2 - 4a √[y2 + (x+c)2] + 2xc 4a √[y2 + (x+c)2] = 4a2 + 4xc a √[y2 + (x+c)2] = a2 + xc

Square both sides again and expand:

a2 (y2 + x2 + 2xc + c2) = a4 + 2a2xc + x2c2 a2y2 + a2x2 + a2c2 = a4 + x2c2 a2x2 - x2c2 + a2y2 = a4 - a2c2

Factor out a2 - c2:

x2 (a2 - c2) + a2y2 = a2(a2 - c2)

Define b2 = a2 - c2

b2x2 + a2y2 = a2b2

Divide to obtain the familiar form:

x2/a2 + y2/b2 = 1

Note that when

x = 0, y = +/- b y = 0, x = +/- a

The squaring method allows the possibility that the simplified equation has solutions that are not valid for the original version. This turns out not to be the case, and Kline deals with this issue in the book.

Also, we might note that since

b2 = a2 - c2

if a and b are fixed, then c is determined. For the figure shown here, b = c and hence a = √2 b.

UPDATE: A virtually identical proof can be found in Wikipedia (here)---should have looked first.