Precalculus Examples

$\frac{x^{2}}{16} - \frac{y^{2}}{25} = 1$

Step 1

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

$\frac{x^{2}}{16} - \frac{y^{2}}{25} = 1$

Step 2

This is the form of a hyperbola. Use this form to determine the values used to find vertices and asymptotes of the hyperbola.

$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$

Step 3

Match the values in this hyperbola to those of the standard form. The variable $h$ represents the x-offset from the origin, $k$ represents the y-offset from origin, $a$ .

$a = 4$

$b = 5$

$k = 0$

$h = 0$

Step 4

Find $c$ , the distance from the center to a focus.

Tap for more steps...

Step 4.1

Find the distance from the center to a focus of the hyperbola by using the following formula.

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

Step 4.2

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(4)}^{2} + {(5)}^{2}}$

Step 4.3

Simplify.

Tap for more steps...

Step 4.3.1

Raise $4$ to the power of $2$ .

$\sqrt{16 + {(5)}^{2}}$

Step 4.3.2

Raise $5$ to the power of $2$ .

$\sqrt{16 + 25}$

Step 4.3.3

Add $16$ and $25$ .

$\sqrt{41}$

Step 5

Find the foci.

Tap for more steps...

Step 5.1

The first focus of a hyperbola can be found by adding $c$ to $h$ .

$(h + c, k)$

Step 5.2

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(\sqrt{41}, 0)$

Step 5.3

The second focus of a hyperbola can be found by subtracting $c$ from $h$ .

$(h - c, k)$

Step 5.4

Substitute the known values of $h$ , $c$ , and $k$ into the formula and simplify.

$(- \sqrt{41}, 0)$

Step 5.5

The foci of a hyperbola follow the form of $(h \pm \sqrt{a^{2} + b^{2}}, k)$ . Hyperbolas have two foci.

$(\sqrt{41}, 0), (- \sqrt{41}, 0)$

Step 6