Precalculus Examples

${(y - 3)}^{2} - 4 {(x - 1)}^{2} = 36$

Step 1

Find the standard form of the hyperbola.

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Step 1.1

Divide each term by $36$ to make the right side equal to one.

$\frac{{(y - 3)}^{2}}{36} - \frac{4 {(x - 1)}^{2}}{36} = \frac{36}{36}$

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 $1$ .

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

Step 2

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

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

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, $a$ .

$a = 6$

$b = 3$

$k = 3$

$h = 1$

Step 4

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

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

Step 5

Simplify to find the first asymptote.

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Step 5.1

Remove parentheses.

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

Step 5.2

Simplify $2 \cdot (x - (1)) + 3$ .

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Multiply $- 1$ by $1$ .

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

Step 5.2.1.2

Apply the distributive property.

$y = 2 x + 2 \cdot - 1 + 3$

Step 5.2.1.3

Multiply $2$ by $- 1$ .

$y = 2 x - 2 + 3$

Step 5.2.2

Add $- 2$ and $3$ .

$y = 2 x + 1$

Step 6

Simplify to find the second asymptote.

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Step 6.1

Remove parentheses.

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

Step 6.2

Simplify $- 2 \cdot (x - (1)) + 3$ .

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Step 6.2.1

Simplify each term.

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Step 6.2.1.1

Multiply $- 1$ by $1$ .

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

Step 6.2.1.2

Apply the distributive property.

$y = - 2 x - 2 \cdot - 1 + 3$

Step 6.2.1.3

Multiply $- 2$ by $- 1$ .

$y = - 2 x + 2 + 3$

Step 6.2.2

Add $2$ and $3$ .

$y = - 2 x + 5$

Step 7

This hyperbola has two asymptotes.

$y = 2 x + 1, y = - 2 x + 5$

Step 8

The asymptotes are $y = 2 x + 1$ and $y = - 2 x + 5$ .

Asymptotes: $y = 2 x + 1, y = - 2 x + 5$

Step 9