Area of an Ellipse

If the semi-axes are equal in length (a = b), the ellipse becomes a circle. In this case, we denote the semi-axes by the radius r (r = a = b), and the formula reduces to the familiar area of a circle: pi times the radius squared.

Area of a Circle = π ⋅ r²

The formula for the area of an ellipse is derived using integration, an advanced Calculus concept not covered in this article. Below, we will walk through several examples of calculating the area of an ellipse.

Examples

Below, we will solve several practice examples to find the area of an ellipse.

Example 1

Find the area of the ellipse given by the equation \(\dfrac{y^2}{64}+\dfrac{x^2}{16}=1\)

Solution: Since the equation is in standard form, we can identify the lengths of the semi-axes by taking the square roots of the denominators beneath the variables:

$$a^2=64→a=\sqrt{64}=8$$

$$b^2=16→b=\sqrt{16}=4$$

With the semi-axes identified, multiply them by pi to find the area:

$$A=\pi\cdot 8\cdot 4=32\pi≈100.53$$

Note: Remember that area is always measured in square units. For instance, if the semi-axes are given in meters (m), the area will be in square meters (m²).

Example 2

Find the area of the ellipse \(\dfrac{(x+3)^2}{36}+\dfrac{(y-4)^2}{25}=1\)

Solution: In this case, the standard equation shows that the ellipse is centered at (-3, 4) rather than the origin. However, translation does not affect the area of the shape.

Find the lengths of the semi-axes:

$$a^2=36→a=\sqrt{36}=6$$

$$b^2=25→b=\sqrt{25}=5$$

Substitute these values into the area formula:

$$A=\pi \cdot 6\cdot 5=30\pi ≈ 94.25$$

Example 3

Find the area of the ellipse given by \(16x^2+25y^2=100.\)

Solution: Since the equation is not in standard form, we cannot determine the lengths of the semi-axes directly.

Dividing both sides by 100 yields the equivalent standard form:

$$\dfrac{x^2}{25/4}+\dfrac{y^2}{4}=1$$

From here, we find the semi-axes:

$$a^2=\dfrac{25}{4}→a=\sqrt{\dfrac{25}{4}}=\dfrac{5}{2}=2.5$$

$$b^2=4→b=\sqrt{4}=2$$

Using these values, we calculate the area:

$$A=\pi \cdot 2.5\cdot 2=5\pi≈15.71$$

Example 4

Find the area of the ellipse \(2x^2+4x+7y^2-28y-40=0\)

Solution: Here, the equation is given in general form rather than standard form. By completing the square, we can rewrite it in the following equivalent form:

$$\dfrac{(x+1)^2}{35}+\dfrac{(y-2)^2}{10}=1$$

This standard equation shows an ellipse shifted away from the origin, from which the lengths of the semi-axes can be identified directly:

$$a^2=35→a=\sqrt{35}$$

$$b^2=10→b=\sqrt{10}$$

Substitute these values into the area formula:

$$A=\pi \cdot \sqrt{35}\cdot \sqrt{10}≈58.77$$

Therefore, the area enclosed by the ellipse is approximately 58.77 square units.