We will derive the equation of the ellipse from its definition. The ellipse is the locus of all points whose sum of the distances to two foci is contant. The equation of the ellipse is \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\).

We will assume we already know that this sum is equal to 2a. We could let it equal some constant d but that is the same as letting it equal 2a and assuming we do not yet know what the a represents.

We will place our foci at (–c, 0) and (c, 0) so that the major axis of the ellipse is the x-axis. We can do a similar exercise if the major axis is the y-axis. Figure 1 below depicts the parts of the ellipse.

Figure 1: The parts of an ellipse (\(|a| > |c|\))

From the definition, we have the following equation and the steps that follow:

(i) \(\sqrt{(x-c)^2+y^2} + \sqrt{(x+c)^2+y^2} = 2a\)

(ii) \(2a - \sqrt{(x-c)^2+y^2} = \sqrt{(x+c)^2+y^2}\)   (Rearrange a bit)

(iii) \(4a^2 - 4a\sqrt{(x-c)^2+y^2} + (x-c)^2 + \cancel{y^2} = (x+c)^2 + \cancel{y^2}\)   (Square both sides)

(iv) \(4a^2 - 4a\sqrt{(x-c)^2+y^2} + \cancel{x^2} - 2cx + \cancel{c^2} + \cancel{y^2} = \cancel{x^2} + 2cx + \cancel{c^2} + \cancel{y^2}\)

(v) \(4a^2 - 4a\sqrt{(x-c)^2+y^2} - 4cx = 0 \)   (Clean up a bit)

(vi) \(a - \frac{c}{a}x = \sqrt{(x-c)^2+y^2} \)   (Rearrange to isolate the radical again and divide by a)

(vii) \(a^2 - \cancel{2cx} + \frac{c^2}{a^2}x^2 = x^2 - \cancel{2cx} + c^2 +y^2 \)   (Square both sides again)

(viii) \(a^4 + c^2x^2 = a^2x^2 + a^2c^2 + a^2y^2 \)   (Clean up and multiply both sides by a²)

(ix) \((a^2-c^2)x^2 + a^2y^2 = a^2(a^2-c^2) \)   (Collect like terms)

(x) \(b^2x^2 + a^2y^2 = a^2b^2 \)   (Let \(a^2 - c^2 = b^2\))

(xi) \( \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \)   (Divide both sides by \(a^2b^2\))

Equation (xi) represents the equation of an ellipse. As of yet, we only know that 2a represents the constant sum of the ellipse, but do not know that this is the length of the major axis. We also know \(a^2 - c^2 = b^2\), but we do not know that 2b represents the length of the minor axis. However, we can easily determine these.

Note that steps (ii) and onward are the exact same steps as deriving the equation of a hyperbola.

The only way to differentiate an ellipse from a hyperbola is to recognize in step (ix) that a > c and let \(a^2 - c^2 = b^2\) instead of \(c^2 - a^2 = b^2\). In Figure 1, you can see that the foci fall inside of the ellipse because |a| is greater than |c|.

The Distances

The Value a

How do we know the constant sum of an ellipse is 2a? We actually used 2a as our constant sum to derive the equation. However, we can establish that again by finding the legnth of the major axis. First, let’s find the x-intercepts because those are easy to find. We let y equal 0 in the ellipse equation to find the x-intercepts.

(i) \( \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 \)

(ii) \( \frac{x^2}{a^2}+0=1 \)

(iii) \( x^2 = a^2 \)

(iv) \( x = \pm a \)

Therefore, the two x-intercepts are (–a, 0) and (a, 0). These two points are the farthest points on the ellipse that make up the major axis which divides the ellipse in half horizontally. And because they are points on the ellipse, their sum of the distances from the foci should equal the constant sum.

The distance from the (–c, 0) to (a, 0) is a + c. This is the distance from one focus to a point on an ellipse. The distance from (a, 0) to (c, 0) is a - c. This is the distance from the same point on the ellipse to the other focus. The sum between these two distances is \((a + c) + (a - c) = a + c + a - c = 2a \).

The Value b

Now, we know that 2a is the length of the major axis and the constant sum of the ellipse. In step (x), we let \(b^2 = a^2 - c^2\). The value b² ended up as the denominator of y² term. If we let x = 0, we obtain the y-intercepts of the ellipse. The math is trivial and (0, –b) and (0, b) are the two y-intercepts.

Figure 2

In Figure 2, we have a point P at the y-intercpet (0, b). Because the constant sum of the ellipse equal 2a, F2P + PF1 = 2a. But because of symmetry, PF1 equals a. And we also know that OF1 = c because that is the focal distance. Therefore, we have the relationship \(b^2 + c^2 = a^2\) from the right triangle POF1.