Focal Distance of an Ellipse

Below are step-by-step examples of how to calculate the focal distance.

How to Calculate the Focal Distance

As stated above, the distance from the center to each focus is related to the semi-axes a and b through the following formula:

\(c^2=a^2-b^2\), which implies that \(c=\sqrt{a^2-b^2}\)

This equation gives the value of c. To find the focal distance, simply multiply c by 2. Alternatively, you can use the focal distance formula directly:

$$2c=2\sqrt{a^2-b^2}$$

If the foci F₁(x₁, y₁) and F₂(x₂, y₂) are known, the distance formula can be used to find the focal distance:

$$2c=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Step-by-Step Examples

Below are several worked examples on finding the focal distance of an ellipse.

Example 1

Find the focal distance of the ellipse \(\dfrac{x^2}{4}+\dfrac{y^2}{25}=1\).

Solution: The problem gives the equation in standard form, where the squares of the semi-axes correspond to the denominators of each variable:

$$a^2=25$$

$$b^2=4$$

Using these values, we calculate the distance from the center to each focus:

$$c=\sqrt{a^2-b^2}$$

$$c=\sqrt{25-4}$$

$$c=\sqrt{21}$$

Now, multiply by 2 to find the focal distance:

$$2c=2\cdot \sqrt{21}≈9.17$$

Therefore, the focal distance of the ellipse is \(2\sqrt{21}\), or approximately 9.17 units.

Example 2

Determine the focal distance of an ellipse whose major axis length is 26 cm and whose minor axis length is 24 cm.

Solution: While the equation is not provided, the lengths of the axes are given, allowing us to find the lengths of the semi-axes:

$$2a=26→a=\dfrac{26}{2}=13$$

$$2b=24→b=\dfrac{24}{2}=12$$

With these values, we calculate the distance c using the formula:

$$c=\sqrt{a^2-b^2}$$

$$c=\sqrt{13^2-12^2}$$

$$c=\sqrt{169-144}$$

$$c=\sqrt{25}$$

$$c=5$$

This gives half of the focal distance. To obtain the full distance between the foci, multiply by 2:

$$2c=2\cdot 5=10$$

Therefore, the focal distance of the ellipse is 10 cm.

Example 3

An ellipse has its foci at coordinates (-2, -1) and (4, -1). Find the focal distance.

Solution: The problem gives the coordinates of the foci. By definition, the focal distance is the distance between them, so we can use the distance formula:

$$2c=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

$$2c=\sqrt{(4-(-2))^2+(-1-(-1))^2}$$

$$2c=\sqrt{6^2+0^2}$$

$$2c=\sqrt{6^2}$$

$$2c=6$$

Therefore, the distance between the foci is 6 units.

Example 4

Find the focal distance of the ellipse whose equation is \(2x^2+4x+7y^2-28y-40=0\).

Solution: The problem gives the general equation of the ellipse, so the values cannot be read directly. However, by completing the square, we can rewrite it in standard form:

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

From this equation, we find:

$$a^2=35$$

$$b^2=10$$

Now, we use the focal distance formula:

$$2c=2\sqrt{a^2-b^2}$$

$$2c=2\sqrt{35-10}$$

$$2c=2\sqrt{25}$$

$$2c=2\cdot 5$$

$$2c=10$$

Therefore, the focal distance of this ellipse is 10 units.