Precalculus Examples

$\frac{y^{2}}{16} - \frac{x^{2}}{36} = 1$

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 $1$ .

$\frac{y^{2}}{16} - \frac{x^{2}}{36} = 1$

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.

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

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, $a$ .

$a = 4$

$b = 6$

$k = 0$

$h = 0$

Step 4

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

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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.

$\sqrt{{(4)}^{2} + {(6)}^{2}}$

Step 4.3

Simplify.

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Step 4.3.1

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

$\sqrt{16 + {(6)}^{2}}$

Step 4.3.2

Raise $6$ to the power of $2$ .

$\sqrt{16 + 36}$

Step 4.3.3

Add $16$ and $36$ .

$\sqrt{52}$

Step 4.3.4

Rewrite $52$ as $2^{2} \cdot 13$ .

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Step 4.3.4.1

Factor $4$ out of $52$ .

$\sqrt{4 (13)}$

Step 4.3.4.2

Rewrite $4$ as $2^{2}$ .

$\sqrt{2^{2} \cdot 13}$

Step 4.3.5

Pull terms out from under the radical.

$2 \sqrt{13}$

Step 5

Find the foci.

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Step 5.1

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

$(h, k + c)$

Step 5.2

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

$(0, 2 \sqrt{13})$

Step 5.3

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

$(h, k - c)$

Step 5.4

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

$(0, - 2 \sqrt{13})$

Step 5.5

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

$(0, 2 \sqrt{13}), (0, - 2 \sqrt{13})$

Step 6