Algebra Examples

$576 x^{2} - 16 y^{2} = 144$

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.

$\frac{576 x^{2}}{144} - \frac{16 y^{2}}{144} = \frac{144}{144}$

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 $1$ .

$\frac{x^{2}}{\frac{1}{4}} - \frac{y^{2}}{9} = 1$

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, $a$ .

$a = \frac{1}{2}$

$b = 3$

$k = 0$

$h = 0$

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.

$\sqrt{{(\frac{1}{2})}^{2} + {(3)}^{2}}$

Step 4.3

Simplify.

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Step 4.3.1

Apply the product rule to $\frac{1}{2}$ .

$\sqrt{\frac{1^{2}}{2^{2}} + {(3)}^{2}}$

Step 4.3.2

One to any power is one.

$\sqrt{\frac{1}{2^{2}} + {(3)}^{2}}$

Step 4.3.3

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

$\sqrt{\frac{1}{4} + {(3)}^{2}}$

Step 4.3.4

Raise $3$ to the power of $2$ .

$\sqrt{\frac{1}{4} + 9}$

Step 4.3.5

To write $9$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\sqrt{\frac{1}{4} + 9 \cdot \frac{4}{4}}$

Step 4.3.6

Combine $9$ and $\frac{4}{4}$ .

$\sqrt{\frac{1}{4} + \frac{9 \cdot 4}{4}}$

Step 4.3.7

Combine the numerators over the common denominator.

$\sqrt{\frac{1 + 9 \cdot 4}{4}}$

Step 4.3.8

Simplify the numerator.

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Step 4.3.8.1

Multiply $9$ by $4$ .

$\sqrt{\frac{1 + 36}{4}}$

Step 4.3.8.2

Add $1$ and $36$ .

$\sqrt{\frac{37}{4}}$

Step 4.3.9

Rewrite $\sqrt{\frac{37}{4}}$ as $\frac{\sqrt{37}}{\sqrt{4}}$ .

$\frac{\sqrt{37}}{\sqrt{4}}$

Step 4.3.10

Simplify the denominator.

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Step 4.3.10.1

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

$\frac{\sqrt{37}}{\sqrt{2^{2}}}$

Step 4.3.10.2

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

$\frac{\sqrt{37}}{2}$

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 $h$ .

$(h + c, k)$

Step 5.2

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

$(\frac{\sqrt{37}}{2}, 0)$

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.

$(- \frac{\sqrt{37}}{2}, 0)$

Step 5.5

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

$(\frac{\sqrt{37}}{2}, 0), (- \frac{\sqrt{37}}{2}, 0)$

Step 6