Derivation of the Ellipse Equation

August 20, 2024

Definition of an Ellipse

An ellipse is the locus of points on a plane such that the sum of the distances from two distinct fixed points (foci) is constant.

Derivation of the Ellipse Equation

Although it involves somewhat tedious calculations, we derive the equation of an ellipse.

We will perform the algebraic manipulations as carefully as possible.

Consider two points $(-c, 0)$ and $(c, 0)$.

Any point $(x, y)$ on the ellipse has a constant sum of distances from these two points.

Let this sum of distances be $L$. Then,

$$ \sqrt{(x + c)^2 + y^2} + \sqrt{(x - c)^2 + y^2} = L $$

Rearrange the second term to the right side:

$$ \sqrt{(x + c)^2 + y^2} = L - \sqrt{(x - c)^2 + y^2} $$

Square both sides:

$$ (x + c)^2 + y^2 = \{L - \sqrt{(x - c)^2 + y^2}\}^2 $$

Expand the right side:

$$ (x + c)^2 + y^2 = L^2 - 2 L \sqrt{(x - c)^2 + y^2} + \{(x - c)^2 + y^2\} $$

Rearrange the final term on the right side to the left side:

$$ \{(x + c)^2 + y^2\} - \{(x - c)^2 + y^2\} = L^2 - 2 L \sqrt{(x - c)^2 + y^2} $$

Expand the left side:

$$ (x^2 + 2cx + c^2 + y^2) - (x^2 - 2cx + c^2 + y^2) = L^2 - 2 L \sqrt{(x - c)^2 + y^2} $$

Simplify the left side:

$$ 4cx = L^2 - 2 L \sqrt{(x - c)^2 + y^2} $$

Rearrange $L^2$ to the left side:

$$ 4cx - L^2 = -2 L \sqrt{(x - c)^2 + y^2} $$

Square both sides:

$$ (4cx - L^2)^2 = 4 L^2 \{(x - c)^2 + y^2\} $$

Expand both sides:

$$ 16c^2x^2 -8L^2cx + L^4 = 4 L^2 x^2 - 8 L^2 cx + 4L^2c^2 + 4L^2y^2 $$

Add $8 L^2 cx$ to both sides:

$$ 16c^2x^2 + L^4 = 4 L^2 x^2 + 4L^2c^2 + 4L^2y^2 $$

Rearrange the terms involving $x$ and $𝑦$ to the left side, and the constants to the right side:

$$ 16c^2x^2 - 4 L^2 x^2 - 4L^2 y^2 = 4L^2c^2 - L^4 $$

Combine $x^2$ terms on the left:

$$ 4(4c^2 - L^2)x^2 - 4L^2 y^2 = 4L^2c^2 - L^4 $$

Combine the right side terms with $L^2$:

$$ 4(4c^2 - L^2)x^2 - 4L^2 y^2 = L^2 (4c^2 - L^2) $$

Devide both sides by $L^2 (4c^2 - L^2)$:

$$ \frac{4x^2}{L^2} - \frac{4y^2}{4c^2 - L^2} = 1 $$

Since $L > 2c$, we have $L^2 - 4c^2 > 0$. Rearranging the terms gives:

$$ \frac{4x^2}{L^2} + \frac{4y^2}{L^2 - 4c^2} = 1 $$

Finally, by dividing both the numerator and the denominator on the left side by 4, we obtain:

$$ \frac{x^2}{(\frac{L}{2})^2} + \frac{y^2}{(\frac{L}{2})^2 - c^2} = 1 $$

Letting:

$$ a = \frac{L}{2} $$

we get:

$$ \frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1 $$

Further, letting:

$$ b^2 = a^2 - c^2 $$

we get:

$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$

This is known as the standard form of the ellipse equation.