Pre-Algebra Examples

$4 x^{2} - 36 y^{2} = 144$

Step 1

Find the standard form of the hyperbola.

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Step 1.1

Divide each term by $144$ to make the right side equal to one.

$\frac{4 x^{2}}{144} - \frac{36 y^{2}}{144} = \frac{144}{144}$

Step 1.2

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 $1$ .

$\frac{x^{2}}{36} - \frac{y^{2}}{4} = 1$

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, $a$ .

$a = 6$

$b = 2$

$k = 0$

$h = 0$

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.

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

Step 5.3

Simplify.

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Step 5.3.1

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

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

Step 5.3.2

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

$\sqrt{36 + 4}$

Step 5.3.3

Add $36$ and $4$ .

$\sqrt{40}$

Step 5.3.4

Rewrite $40$ as $2^{2} \cdot 10$ .

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Step 5.3.4.1

Factor $4$ out of $40$ .

$\sqrt{4 (10)}$

Step 5.3.4.2

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

$\sqrt{2^{2} \cdot 10}$

Step 5.3.5

Pull terms out from under the radical.

$2 \sqrt{10}$

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.

$(6, 0)$

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.

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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.

$(2 \sqrt{10}, 0)$

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.

$(- 2 \sqrt{10}, 0)$

Step 7.5

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

$(2 \sqrt{10}, 0), (- 2 \sqrt{10}, 0)$

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.

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

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 $6$ to the power of $2$ .

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

Step 8.3.1.2

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

$\frac{\sqrt{36 + 4}}{6}$

Step 8.3.1.3

Add $36$ and $4$ .

$\frac{\sqrt{40}}{6}$

Step 8.3.1.4

Rewrite $40$ as $2^{2} \cdot 10$ .

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Step 8.3.1.4.1

Factor $4$ out of $40$ .

$\frac{\sqrt{4 (10)}}{6}$

Step 8.3.1.4.2

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

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

Step 8.3.1.5

Pull terms out from under the radical.

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

Step 8.3.2

Cancel the common factor of $2$ and $6$ .

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Step 8.3.2.1

Factor $2$ out of $2 \sqrt{10}$ .

$\frac{2 (\sqrt{10})}{6}$

Step 8.3.2.2

Cancel the common factors.

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Step 8.3.2.2.1

Factor $2$ out of $6$ .

$\frac{2 \sqrt{10}}{2 \cdot 3}$

Step 8.3.2.2.2

Cancel the common factor.

$\frac{2 \sqrt{10}}{2 \cdot 3}$

Step 8.3.2.2.3

Rewrite the expression.

$\frac{\sqrt{10}}{3}$

Step 9

Find the focal parameter.

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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.

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

Step 9.3

Simplify.

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Step 9.3.1

Cancel the common factor of $2^{2}$ and $2$ .

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Step 9.3.1.1

Factor $2$ out of $2^{2}$ .

$\frac{2 \cdot 2}{2 \sqrt{10}}$

Step 9.3.1.2

Cancel the common factors.

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Step 9.3.1.2.1

Factor $2$ out of $2 \sqrt{10}$ .

$\frac{2 \cdot 2}{2 (\sqrt{10})}$

Step 9.3.1.2.2

Cancel the common factor.

$\frac{2 \cdot 2}{2 \sqrt{10}}$

Step 9.3.1.2.3

Rewrite the expression.

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

Step 9.3.2

Multiply $\frac{2}{\sqrt{10}}$ by $\frac{\sqrt{10}}{\sqrt{10}}$ .

$\frac{2}{\sqrt{10}} \cdot \frac{\sqrt{10}}{\sqrt{10}}$

Step 9.3.3

Combine and simplify the denominator.

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Step 9.3.3.1

Multiply $\frac{2}{\sqrt{10}}$ by $\frac{\sqrt{10}}{\sqrt{10}}$ .

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

Step 9.3.3.2

Raise $\sqrt{10}$ to the power of $1$ .

$\frac{2 \sqrt{10}}{{\sqrt{10}}^{1} \sqrt{10}}$

Step 9.3.3.3

Raise $\sqrt{10}$ to the power of $1$ .

$\frac{2 \sqrt{10}}{{\sqrt{10}}^{1} {\sqrt{10}}^{1}}$

Step 9.3.3.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{2 \sqrt{10}}{{\sqrt{10}}^{1 + 1}}$

Step 9.3.3.5

Add $1$ and $1$ .

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

Step 9.3.3.6

Rewrite ${\sqrt{10}}^{2}$ as $10$ .

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Step 9.3.3.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{10}$ as $10^{\frac{1}{2}}$ .

$\frac{2 \sqrt{10}}{{(10^{\frac{1}{2}})}^{2}}$

Step 9.3.3.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{2 \sqrt{10}}{10^{\frac{1}{2} \cdot 2}}$

Step 9.3.3.6.3

Combine $\frac{1}{2}$ and $2$ .

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

Step 9.3.3.6.4

Cancel the common factor of $2$ .

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Step 9.3.3.6.4.1

Cancel the common factor.

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

Step 9.3.3.6.4.2

Rewrite the expression.

$\frac{2 \sqrt{10}}{10^{1}}$

Step 9.3.3.6.5

Evaluate the exponent.

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

Step 9.3.4

Cancel the common factor of $2$ and $10$ .

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Step 9.3.4.1

Factor $2$ out of $2 \sqrt{10}$ .

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

Step 9.3.4.2

Cancel the common factors.

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Step 9.3.4.2.1

Factor $2$ out of $10$ .

$\frac{2 \sqrt{10}}{2 \cdot 5}$

Step 9.3.4.2.2

Cancel the common factor.

$\frac{2 \sqrt{10}}{2 \cdot 5}$

Step 9.3.4.2.3

Rewrite the expression.

$\frac{\sqrt{10}}{5}$

Step 10

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

$y = \pm \frac{1}{3} x + 0$

Step 11

Simplify $\frac{1}{3} x + 0$ .

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Step 11.1

Add $\frac{1}{3} x$ and $0$ .

$y = \frac{1}{3} x$

Step 11.2

Combine $\frac{1}{3}$ and $x$ .

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

Step 12

Simplify $- \frac{1}{3} x + 0$ .

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Step 12.1

Add $- \frac{1}{3} x$ and $0$ .

$y = - \frac{1}{3} x$

Step 12.2

Combine $x$ and $\frac{1}{3}$ .

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

Step 13

This hyperbola has two asymptotes.

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

Step 14

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

Center: $(0, 0)$

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

Foci: $(2 \sqrt{10}, 0), (- 2 \sqrt{10}, 0)$

Eccentricity: $\frac{\sqrt{10}}{3}$

Focal Parameter: $\frac{\sqrt{10}}{5}$

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

Step 15