Focal Distance of a Hyperbola

How to Calculate the Focal Distance

As noted earlier, the distance from the center to each focus (c) is related to the semi-transverse axis (a) and semi-conjugate axis (b) by the following formula:

c² = a² + b²

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

This expression allows us to find the value of c from a and b, which can be identified from the standard equation of the hyperbola. Multiplying c by two gives the focal distance.

If the coordinates of the foci F₁(x₁, y₁) and F₂(x₂, y₂) are known, we can use the distance formula between two points to find the focal distance:

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

Step-by-Step Examples

Example 1

Find the focal distance of the hyperbola \(\dfrac{x^2}{4}-\dfrac{y^2}{9}=1\)

Solution: From the equation, we observe that the hyperbola has a horizontal transverse axis along the x-axis, as the positive term contains x². The values of a² and b² are the denominators of the quadratic terms: \(a^2=4\quad \text{and}\quad b^2=9.\)

To calculate c, we use the formula:

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

Substituting the values of \(a^2\quad \text{and}\quad b^2:\)

$$c=\sqrt{4+9}$$

$$c=\sqrt{13}$$

Therefore, the focal distance of the hyperbola is twice c: \(2c=2\sqrt{13}.\)

Example 2

Given the equation of the hyperbola \(\dfrac{(y-1)^2}{25}-\dfrac{(x+1)^2}{4}=1\), calculate its focal distance.

Solution: We identify that the equation is in standard form for a hyperbola with a vertical transverse axis. The center of the hyperbola is (-1, 1), and the values of a² and b² are:

$$a^2=25$$

$$b^2=4$$

To find c, we use the relationship:

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

Substituting the values of \(a^2\quad \text{and}\quad b^2:\)

$$c=\sqrt{25+4}=\sqrt{29}$$

The focal distance of the hyperbola is twice c; that is:

$$2c=2\sqrt{29}$$

Example 3

Determine the distance between the foci of the hyperbola \(\dfrac{x^2}{12}-\dfrac{(y+2)^2}{36}=1\)

Solution: We identify that the equation is in standard form for a hyperbola with a horizontal transverse axis, with its center at (0, -2), and values:

$$a^2=12$$

$$b^2=36$$

To find c, we use the formula:

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

Substituting the values of \(a^2 \quad \text{and}\quad b^2:\)

$$c=\sqrt{12+36}=\sqrt{48}$$

Simplifying \(\sqrt{48}:\)

$$c=\sqrt{16 \cdot 3}=4\sqrt{3}$$

The focal distance of the hyperbola is twice c; that is:

$$2c=2 \cdot 4\sqrt{3}=8\sqrt{3}$$

Example 4

Find the focal distance of the hyperbola whose equation is \(5x^2-2y^2-10x+8y-13=0\)

Solution: The equation is not in standard form, so we cannot read the lengths of the semi-axes directly. However, we can complete the square to obtain an equivalent form:

$$\dfrac{(x-1)^2}{2}-\dfrac{(y-2)^2}{5}=1$$

First, we identify that the equation represents a hyperbola with a horizontal transverse axis, centered at (1, 2), where:

$$a^2=2\quad \text{and}\quad b^2=5$$

To calculate c, we use the formula:

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

Substituting the values of \(a^2\quad \text{and}\quad b^2:\)

$$c=\sqrt{2+5}=\sqrt{7}$$

The focal distance of the hyperbola is twice c; that is:

$$2c=2\sqrt{7}$$