Precalculus Examples

\frac{{(x - 2)}^{2}}{36} - \frac{{(y - 3)}^{2}}{9} = 1

$\frac{{(x - 2)}^{2}}{36} - \frac{{(y - 3)}^{2}}{9} = 1$

Step 1

Simplify each term in the equation in order to set the right side equal to

1

$1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be

1

$1$ .

\frac{{(x - 2)}^{2}}{36} - \frac{{(y - 3)}^{2}}{9} = 1

$\frac{{(x - 2)}^{2}}{36} - \frac{{(y - 3)}^{2}}{9} = 1$

Step 2

This is the form of a hyperbola. Use this form to determine the values used to find the asymptotes of the hyperbola.

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

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

Step 3

Match the values in this hyperbola to those of the standard form. The variable

h

$h$ represents the x-offset from the origin,

k

$k$ represents the y-offset from origin,

a

$a$ .

a = 6

$a = 6$

b = 3

$b = 3$

k = 3

$k = 3$

h = 2

$h = 2$

Step 4

The asymptotes follow the form

y = \pm \frac{b (x - h)}{a} + k

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

y = \pm \frac{1}{2} \cdot (x - (2)) + 3

$y = \pm \frac{1}{2} \cdot (x - (2)) + 3$

Step 5

Simplify to find the first asymptote.

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Step 5.1

Remove parentheses.

y = \frac{1}{2} \cdot (x - (2)) + 3

$y = \frac{1}{2} \cdot (x - (2)) + 3$

Step 5.2

Simplify

\frac{1}{2} \cdot (x - (2)) + 3

$\frac{1}{2} \cdot (x - (2)) + 3$ .

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Multiply

- 1

$- 1$ by

2

$2$ .

y = \frac{1}{2} \cdot (x - 2) + 3

$y = \frac{1}{2} \cdot (x - 2) + 3$

Step 5.2.1.2

Apply the distributive property.

y = \frac{1}{2} x + \frac{1}{2} \cdot - 2 + 3

$y = \frac{1}{2} x + \frac{1}{2} \cdot - 2 + 3$

Step 5.2.1.3

Combine

\frac{1}{2}

$\frac{1}{2}$ and

x

$x$ .

y = \frac{x}{2} + \frac{1}{2} \cdot - 2 + 3

$y = \frac{x}{2} + \frac{1}{2} \cdot - 2 + 3$

Step 5.2.1.4

Cancel the common factor of

2

$2$ .

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Step 5.2.1.4.1

Factor

2

$2$ out of

- 2

$- 2$ .

y = \frac{x}{2} + \frac{1}{2} \cdot (2 (- 1)) + 3

$y = \frac{x}{2} + \frac{1}{2} \cdot (2 (- 1)) + 3$

Step 5.2.1.4.2

Cancel the common factor.

$y = \frac{x}{2} + \frac{1}{2} \cdot (2 \cdot - 1) + 3$

Step 5.2.1.4.3

Rewrite the expression.

$y = \frac{x}{2} - 1 + 3$

Step 5.2.2

Add

$- 1$ and

$3$ .

$y = \frac{x}{2} + 2$

Step 6

Simplify to find the second asymptote.

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Step 6.1

Remove parentheses.

$y = - \frac{1}{2} \cdot (x - (2)) + 3$

Step 6.2

Simplify

$- \frac{1}{2} \cdot (x - (2)) + 3$ .

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Multiply

$- 1$ by

$2$ .

$y = - \frac{1}{2} \cdot (x - 2) + 3$

Step 6.2.1.2

Apply the distributive property.

$y = - \frac{1}{2} x - \frac{1}{2} \cdot - 2 + 3$

Step 6.2.1.3

Combine

$x$ and

$\frac{1}{2}$ .

$y = - \frac{x}{2} - \frac{1}{2} \cdot - 2 + 3$

Step 6.2.1.4

Cancel the common factor of

$2$ .

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Step 6.2.1.4.1

Move the leading negative in

$- \frac{1}{2}$ into the numerator.

$y = - \frac{x}{2} + \frac{- 1}{2} \cdot - 2 + 3$

Step 6.2.1.4.2

Factor

$2$ out of

$- 2$ .

$y = - \frac{x}{2} + \frac{- 1}{2} \cdot (2 (- 1)) + 3$

Step 6.2.1.4.3

Cancel the common factor.

$y = - \frac{x}{2} + \frac{- 1}{2} \cdot (2 \cdot - 1) + 3$

Step 6.2.1.4.4

Rewrite the expression.

$y = - \frac{x}{2} - 1 \cdot - 1 + 3$

Step 6.2.1.5

Multiply

$- 1$ by

$- 1$ .

$y = - \frac{x}{2} + 1 + 3$

Step 6.2.2

Add

$1$ and

$3$ .

$y = - \frac{x}{2} + 4$

Step 7

This hyperbola has two asymptotes.

$y = \frac{x}{2} + 2, y = - \frac{x}{2} + 4$

Step 8

The asymptotes are

$y = \frac{x}{2} + 2$ and

$y = - \frac{x}{2} + 4$ .

Asymptotes:

$y = \frac{x}{2} + 2, y = - \frac{x}{2} + 4$

Step 9