Precalculus Examples

$\frac{x^{2}}{4} - \frac{y^{2}}{4} = 1$

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 $1$ .

$\frac{x^{2}}{4} - \frac{y^{2}}{4} = 1$

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, $a$ .

$a = 2$

$b = 2$

$k = 0$

$h = 0$

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.

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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.

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

Step 5.3

Simplify.

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Step 5.3.1

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

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

Step 5.3.2

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

$\sqrt{4 + 4}$

Step 5.3.3

Add $4$ and $4$ .

$\sqrt{8}$

Step 5.3.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 5.3.4.1

Factor $4$ out of $8$ .

$\sqrt{4 (2)}$

Step 5.3.4.2

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

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

Step 5.3.5

Pull terms out from under the radical.

$2 \sqrt{2}$

Step 6

Find the vertices.

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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.

$(2, 0)$

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.

$(- 2, 0)$

Step 6.5

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

$(2, 0), (- 2, 0)$

Step 7

Find the foci.

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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.

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

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.

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

Step 7.5

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

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

Step 8

Find the eccentricity.

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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.

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

Step 8.3

Simplify.

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Step 8.3.1

Simplify the numerator.

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Step 8.3.1.1

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

$\frac{\sqrt{4 + 2^{2}}}{2}$

Step 8.3.1.2

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

$\frac{\sqrt{4 + 4}}{2}$

Step 8.3.1.3

Add $4$ and $4$ .

$\frac{\sqrt{8}}{2}$

Step 8.3.1.4

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 8.3.1.4.1

Factor $4$ out of $8$ .

$\frac{\sqrt{4 (2)}}{2}$

Step 8.3.1.4.2

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

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

Step 8.3.1.5

Pull terms out from under the radical.

$\frac{2 \sqrt{2}}{2}$

Step 8.3.2

Cancel the common factor of $2$ .

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Step 8.3.2.1

Cancel the common factor.

$\frac{2 \sqrt{2}}{2}$

Step 8.3.2.2

Divide $\sqrt{2}$ by $1$ .

$\sqrt{2}$

Step 9

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

$y = \pm 1 \cdot x + 0$

Step 10

Simplify $1 \cdot x + 0$ .

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Step 10.1

Add $1 \cdot x$ and $0$ .

$y = 1 \cdot x$

Step 10.2

Multiply $x$ by $1$ .

$y = x$

Step 11

Simplify $- 1 \cdot x + 0$ .

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Step 11.1

Add $- 1 \cdot x$ and $0$ .

$y = - 1 \cdot x$

Step 11.2

Rewrite $- 1 x$ as $- x$ .

$y = - x$

Step 12

This hyperbola has two asymptotes.

$y = x, y = - x$

Step 13

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

Center: $(0, 0)$

Vertices: $(2, 0), (- 2, 0)$

Foci: $(2 \sqrt{2}, 0), (- 2 \sqrt{2}, 0)$

Eccentricity: $\sqrt{2}$

Focal Parameter: $\sqrt{2}$

Asymptotes: $y = x$ , $y = - x$

Step 14