Home » Ellipses

Tag: Ellipses

Use integrals to establish the formula for the area of an ellipse

Denoting the unit circle, x^2 + y^2 = 1, by C we define an ellipse E to be the set of points

    \[ E = \{ (ax, by) \mid (x,y) \in C,  a > 0,  b > 0 \}. \]

  1. Show that the points on E satisfy the equation:

        \[ \left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1. \]

  2. Prove that the area of E is measurable and that

        \[ a(E) = \pi a b. \]

  1. Proof. If (x,y) is a point on E then \left( \frac{x}{a}, \frac{y}{b} \right) is a point on C (since all points of E are obtained by taking a point of C and multiplying the x-coordinate by a and the y-coordinate by b). By definition of C, we must then have,

        \[ \left( \frac{x}{a} \right)^2 + \left(\frac{y}{b} \right)^2 = 1. \qquad \blacksquare \]

  2. Proof. From part (a) we know E is the set of points (x,y) such that \left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1. But, this implies,

        \[ y = b \cdot \sqrt{ 1 - \left( \frac{x}{a} \right)^2}, \qquad \text{or} \qquad y = -b \cdot \sqrt{1 - \left( \frac{x}{a} \right)^2}. \]

    Hence, the area of E is the area enclosed from -a to a by the graphs of

        \[ g(x) = b \cdot \sqrt{1 - \left( \frac{x}{a} \right)^2} \qquad \text{and} \qquad f(x) = -b \cdot \sqrt{1 - \left( \frac{x}{a} \right)^2}. \]

    To show this region is measurable and has area \pi a b we begin with this identity (from Apostol, 2.4 Exercise 17),

        \begin{align*}  \pi = 2 \int_{-1}^1 \sqrt{1-x^2} && \implies && \pi b &= 2 \int_{-1}^1 b \sqrt{1-x^2} \, dx \\ && \implies && \pi a b &= 2a \int_{-1}^1 b \sqrt{1-x^2} \, dx \\ && \implies && \pi a b &= 2 \int_{-a}^a b \sqrt{1 - \left( \frac{x}{a} \right)^2} \, dx \\ && \implies && \pi ab &= \int_{-a}^a \left( b \sqrt{1 - \left( \frac{x}{a} \right)} - \left(-b \sqrt{1 - \left( \frac{x}{a} \right)^2} \right)\right) \, dx. \end{align*}

    In the second to last line we have used the expansion/contraction of the interval of integration. Hence, we know the integral from -a to a of g(x) - f(x) exists and has value \pi a b. Thus, E is measurable and a(E) = \pi a b. \qquad \blacksquare