Precalculus Examples

$\frac{x^{2}}{36} - \frac{y^{2}}{64} = 1$

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 $1$ .

$\frac{x^{2}}{36} - \frac{y^{2}}{64} = 1$

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, $a$ .

$a = 6$

$b = 8$

$k = 0$

$h = 0$

Step 4

The center of a hyperbola follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(0, 0)$

Step 5

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

Tap for more steps...

Step 5.1

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

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

Step 5.2

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

$\sqrt{{(6)}^{2} + {(8)}^{2}}$

Step 5.3

Simplify.

Tap for more steps...

Step 5.3.1

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

$\sqrt{36 + {(8)}^{2}}$

Step 5.3.2

Raise $8$ to the power of $2$ .

$\sqrt{36 + 64}$

Step 5.3.3

Add $36$ and $64$ .

$\sqrt{100}$

Step 5.3.4

Rewrite $100$ as $10^{2}$ .

$\sqrt{10^{2}}$

Step 5.3.5

Pull terms out from under the radical, assuming positive real numbers.

$10$

Step 6

Find the vertices.

Tap for more steps...

Step 6.1

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

$(h + a, k)$

Step 6.2

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

$(6, 0)$

Step 6.3

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

$(h - a, k)$

Step 6.4

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

$(- 6, 0)$

Step 6.5

The vertices of a hyperbola follow the form of $(h \pm a, k)$ . Hyperbolas have two vertices.

$(6, 0), (- 6, 0)$

Step 7

Find the foci.

Tap for more steps...

Step 7.1

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

$(h + c, k)$

Step 7.2

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

$(10, 0)$

Step 7.3

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

$(h - c, k)$

Step 7.4

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

$(- 10, 0)$

Step 7.5

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

$(10, 0), (- 10, 0)$

Step 8

Find the eccentricity.

Tap for more steps...

Step 8.1

Find the eccentricity by using the following formula.

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

Step 8.2

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

$\frac{\sqrt{{(6)}^{2} + {(8)}^{2}}}{6}$

Step 8.3

Simplify.

Tap for more steps...

Step 8.3.1

Simplify the numerator.

Tap for more steps...

Step 8.3.1.1

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

$\frac{\sqrt{36 + 8^{2}}}{6}$

Step 8.3.1.2

Raise $8$ to the power of $2$ .

$\frac{\sqrt{36 + 64}}{6}$

Step 8.3.1.3

Add $36$ and $64$ .

$\frac{\sqrt{100}}{6}$

Step 8.3.1.4

Rewrite $100$ as $10^{2}$ .

$\frac{\sqrt{10^{2}}}{6}$

Step 8.3.1.5

Pull terms out from under the radical, assuming positive real numbers.

$\frac{10}{6}$

Step 8.3.2

Cancel the common factor of $10$ and $6$ .

Tap for more steps...

Step 8.3.2.1

Factor $2$ out of $10$ .

$\frac{2 (5)}{6}$

Step 8.3.2.2

Cancel the common factors.

Tap for more steps...

Step 8.3.2.2.1

Factor $2$ out of $6$ .

$\frac{2 \cdot 5}{2 \cdot 3}$

Step 8.3.2.2.2

Cancel the common factor.

$\frac{2 \cdot 5}{2 \cdot 3}$

Step 8.3.2.2.3

Rewrite the expression.

$\frac{5}{3}$

Step 9

Find the focal parameter.

Tap for more steps...

Step 9.1

Find the value of the focal parameter of the hyperbola by using the following formula.

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

Step 9.2

Substitute the values of $b$ and $\sqrt{a^{2} + b^{2}}$ in the formula.

$\frac{8^{2}}{10}$

Step 9.3

Simplify.

Tap for more steps...

Step 9.3.1

Raise $8$ to the power of $2$ .

$\frac{64}{10}$

Step 9.3.2

Cancel the common factor of $64$ and $10$ .

Tap for more steps...

Step 9.3.2.1

Factor $2$ out of $64$ .

$\frac{2 (32)}{10}$

Step 9.3.2.2

Cancel the common factors.

Tap for more steps...

Step 9.3.2.2.1

Factor $2$ out of $10$ .

$\frac{2 \cdot 32}{2 \cdot 5}$

Step 9.3.2.2.2

Cancel the common factor.

$\frac{2 \cdot 32}{2 \cdot 5}$

Step 9.3.2.2.3

Rewrite the expression.

$\frac{32}{5}$

Step 10

The asymptotes follow the form $y = \pm \frac{b (x - h)}{a} + k$ because this hyperbola opens left and right.

$y = \pm \frac{4}{3} x + 0$

Step 11

Simplify $\frac{4}{3} x + 0$ .

Tap for more steps...

Step 11.1

Add $\frac{4}{3} x$ and $0$ .

$y = \frac{4}{3} x$

Step 11.2

Combine $\frac{4}{3}$ and $x$ .

$y = \frac{4 x}{3}$

Step 12

Simplify $- \frac{4}{3} x + 0$ .

Tap for more steps...

Step 12.1

Add $- \frac{4}{3} x$ and $0$ .

$y = - \frac{4}{3} x$

Step 12.2

Combine $x$ and $\frac{4}{3}$ .

$y = - \frac{x \cdot 4}{3}$

Step 12.3

Move $4$ to the left of $x$ .

$y = - \frac{4 x}{3}$

Step 13

This hyperbola has two asymptotes.

$y = \frac{4 x}{3}, y = - \frac{4 x}{3}$

Step 14

These values represent the important values for graphing and analyzing a hyperbola.

Center: $(0, 0)$

Vertices: $(6, 0), (- 6, 0)$

Foci: $(10, 0), (- 10, 0)$

Eccentricity: $\frac{5}{3}$

Focal Parameter: $\frac{32}{5}$

Asymptotes: $y = \frac{4 x}{3}$ , $y = - \frac{4 x}{3}$

Step 15