Precalculus Examples

$9 x^{2} - y^{2} - 36 x - 6 y + 18 = 0$

Step 1

Find the standard form of the hyperbola.

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Step 1.1

Subtract $18$ from both sides of the equation.

$9 x^{2} - y^{2} - 36 x - 6 y = - 18$

Step 1.2

Complete the square for $9 x^{2} - 36 x$ .

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Step 1.2.1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 9$

$b = - 36$

$c = 0$

Step 1.2.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 1.2.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 1.2.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 36}{2 \cdot 9}$

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

Cancel the common factor of $- 36$ and $2$ .

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Step 1.2.3.2.1.1

Factor $2$ out of $- 36$ .

$d = \frac{2 \cdot - 18}{2 \cdot 9}$

Step 1.2.3.2.1.2

Cancel the common factors.

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Step 1.2.3.2.1.2.1

Factor $2$ out of $2 \cdot 9$ .

$d = \frac{2 \cdot - 18}{2 (9)}$

Step 1.2.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot - 18}{2 \cdot 9}$

Step 1.2.3.2.1.2.3

Rewrite the expression.

$d = \frac{- 18}{9}$

Step 1.2.3.2.2

Cancel the common factor of $- 18$ and $9$ .

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Step 1.2.3.2.2.1

Factor $9$ out of $- 18$ .

$d = \frac{9 \cdot - 2}{9}$

Step 1.2.3.2.2.2

Cancel the common factors.

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Step 1.2.3.2.2.2.1

Factor $9$ out of $9$ .

$d = \frac{9 \cdot - 2}{9 (1)}$

Step 1.2.3.2.2.2.2

Cancel the common factor.

$d = \frac{9 \cdot - 2}{9 \cdot 1}$

Step 1.2.3.2.2.2.3

Rewrite the expression.

$d = \frac{- 2}{1}$

Step 1.2.3.2.2.2.4

Divide $- 2$ by $1$ .

$d = - 2$

Step 1.2.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 1.2.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 36)}^{2}}{4 \cdot 9}$

Step 1.2.4.2

Simplify the right side.

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Step 1.2.4.2.1

Simplify each term.

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Step 1.2.4.2.1.1

Raise $- 36$ to the power of $2$ .

$e = 0 - \frac{1296}{4 \cdot 9}$

Step 1.2.4.2.1.2

Multiply $4$ by $9$ .

$e = 0 - \frac{1296}{36}$

Step 1.2.4.2.1.3

Divide $1296$ by $36$ .

$e = 0 - 1 \cdot 36$

Step 1.2.4.2.1.4

Multiply $- 1$ by $36$ .

$e = 0 - 36$

Step 1.2.4.2.2

Subtract $36$ from $0$ .

$e = - 36$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $9 {(x - 2)}^{2} - 36$ .

$9 {(x - 2)}^{2} - 36$

Step 1.3

Substitute $9 {(x - 2)}^{2} - 36$ for $9 x^{2} - 36 x$ in the equation $9 x^{2} - y^{2} - 36 x - 6 y = - 18$ .

$9 {(x - 2)}^{2} - 36 - y^{2} - 6 y = - 18$

Step 1.4

Move $- 36$ to the right side of the equation by adding $36$ to both sides.

$9 {(x - 2)}^{2} - y^{2} - 6 y = - 18 + 36$

Step 1.5

Complete the square for $- y^{2} - 6 y$ .

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Step 1.5.1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = - 1$

$b = - 6$

$c = 0$

Step 1.5.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 1.5.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 1.5.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 6}{2 \cdot - 1}$

Step 1.5.3.2

Simplify the right side.

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Step 1.5.3.2.1

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

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Step 1.5.3.2.1.1

Factor $2$ out of $- 6$ .

$d = \frac{2 \cdot - 3}{2 \cdot - 1}$

Step 1.5.3.2.1.2

Move the negative one from the denominator of $\frac{- 3}{- 1}$ .

$d = - 1 \cdot - 3$

Step 1.5.3.2.2

Rewrite $- 1 \cdot - 3$ as $- - 3$ .

$d = - - 3$

Step 1.5.3.2.3

Multiply $- 1$ by $- 3$ .

$d = 3$

Step 1.5.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 1.5.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 6)}^{2}}{4 \cdot - 1}$

Step 1.5.4.2

Simplify the right side.

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Step 1.5.4.2.1

Simplify each term.

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Step 1.5.4.2.1.1

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

$e = 0 - \frac{36}{4 \cdot - 1}$

Step 1.5.4.2.1.2

Multiply $4$ by $- 1$ .

$e = 0 - \frac{36}{- 4}$

Step 1.5.4.2.1.3

Divide $36$ by $- 4$ .

$e = 0 - - 9$

Step 1.5.4.2.1.4

Multiply $- 1$ by $- 9$ .

$e = 0 + 9$

Step 1.5.4.2.2

Add $0$ and $9$ .

$e = 9$

Step 1.5.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(y + 3)}^{2} + 9$ .

$- {(y + 3)}^{2} + 9$

Step 1.6

Substitute $- {(y + 3)}^{2} + 9$ for $- y^{2} - 6 y$ in the equation $9 x^{2} - y^{2} - 36 x - 6 y = - 18$ .

$9 {(x - 2)}^{2} - {(y + 3)}^{2} + 9 = - 18 + 36$

Step 1.7

Move $9$ to the right side of the equation by adding $9$ to both sides.

$9 {(x - 2)}^{2} - {(y + 3)}^{2} = - 18 + 36 - 9$

Step 1.8

Simplify $- 18 + 36 - 9$ .

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Step 1.8.1

Add $- 18$ and $36$ .

$9 {(x - 2)}^{2} - {(y + 3)}^{2} = 18 - 9$

Step 1.8.2

Subtract $9$ from $18$ .

$9 {(x - 2)}^{2} - {(y + 3)}^{2} = 9$

Step 1.9

Divide each term by $9$ to make the right side equal to one.

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

Step 1.10

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 $1$ .

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

Step 2

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

$a = 1$

$b = 3$

$k = - 3$

$h = 2$

Step 4

The center of a hyperbola follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(2, - 3)$

Step 5