Precalculus Examples

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

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

$1$ .

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

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.

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

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,

$a$ .

$a = 8$

$b = 6$

$k = 0$

$h = 0$

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.

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

Step 5.3

Simplify.

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Step 5.3.1

Raise

$8$ to the power of

$2$ .

$\sqrt{64 + {(6)}^{2}}$

Step 5.3.2

Raise

$6$ to the power of

$2$ .

$\sqrt{64 + 36}$

Step 5.3.3

Add

$64$ and

$36$ .

$\sqrt{100}$

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.

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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.

$(8, 0)$

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.

$(- 8, 0)$

Step 6.5

The vertices of a hyperbola follow the form of

$(h \pm a, k)$ . Hyperbolas have two vertices.

$(8, 0), (- 8, 0)$

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.

$(10, 0)$

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.

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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.

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

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

$8$ to the power of

$2$ .

$\frac{\sqrt{64 + 6^{2}}}{8}$

Step 8.3.1.2

Raise

$6$ to the power of

$2$ .

$\frac{\sqrt{64 + 36}}{8}$

Step 8.3.1.3

Add

$64$ and

$36$ .

$\frac{\sqrt{100}}{8}$

Step 8.3.1.4

Rewrite

$100$ as

$10^{2}$ .

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

Step 8.3.1.5

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

$\frac{10}{8}$

Step 8.3.2

Cancel the common factor of

$10$ and

$8$ .

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Step 8.3.2.1

Factor

$2$ out of

$10$ .

$\frac{2 (5)}{8}$

Step 8.3.2.2

Cancel the common factors.

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Step 8.3.2.2.1

Factor

$2$ out of

$8$ .

$\frac{2 \cdot 5}{2 \cdot 4}$

Step 8.3.2.2.2

Cancel the common factor.

$\frac{2 \cdot 5}{2 \cdot 4}$

Step 8.3.2.2.3

Rewrite the expression.

$\frac{5}{4}$

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.

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

Step 9.3

Simplify.

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Step 9.3.1

Raise

$6$ to the power of

$2$ .

$\frac{36}{10}$

Step 9.3.2

Cancel the common factor of

$36$ and

$10$ .

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Step 9.3.2.1

Factor

$2$ out of

$36$ .

$\frac{2 (18)}{10}$

Step 9.3.2.2

Cancel the common factors.

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Step 9.3.2.2.1

Factor

$2$ out of

$10$ .

$\frac{2 \cdot 18}{2 \cdot 5}$

Step 9.3.2.2.2

Cancel the common factor.

$\frac{2 \cdot 18}{2 \cdot 5}$

Step 9.3.2.2.3

Rewrite the expression.

$\frac{18}{5}$

Step 10

The asymptotes follow the form

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

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

Step 11

Simplify

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

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Step 11.1

Add

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

$0$ .

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

Step 11.2

Combine

$\frac{3}{4}$ and

$x$ .

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

Step 12

Simplify

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

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Step 12.1

Add

$- \frac{3}{4} x$ and

$0$ .

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

Step 12.2

Combine

$x$ and

$\frac{3}{4}$ .

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

Step 12.3

Move

$3$ to the left of

$x$ .

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

Step 13

This hyperbola has two asymptotes.

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

Step 14

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

Center:

$(0, 0)$

Vertices:

$(8, 0), (- 8, 0)$

Foci:

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

Eccentricity:

$\frac{5}{4}$

Focal Parameter:

$\frac{18}{5}$

Asymptotes:

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

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

Step 15