Precalculus Examples

$x^{2} - y^{2} - 6 x - 4 y - 11 = 0$

Step 1

Find the standard form of the hyperbola.

Tap for more steps...

Step 1.1

Add $11$ to both sides of the equation.

$x^{2} - y^{2} - 6 x - 4 y = 11$

Step 1.2

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

Tap for more steps...

Step 1.2.1

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 1.2.2

Consider the vertex form of a parabola.

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

Step 1.2.3

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

Tap for more steps...

Step 1.2.3.1

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

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

Step 1.2.3.2

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

Tap for more steps...

Step 1.2.3.2.1

Factor $2$ out of $- 6$ .

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

Step 1.2.3.2.2

Cancel the common factors.

Tap for more steps...

Step 1.2.3.2.2.1

Factor $2$ out of $2 \cdot 1$ .

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

Step 1.2.3.2.2.2

Cancel the common factor.

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

Step 1.2.3.2.2.3

Rewrite the expression.

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

Step 1.2.3.2.2.4

Divide $- 3$ by $1$ .

$d = - 3$

Step 1.2.4

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

Tap for more steps...

Step 1.2.4.1

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

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

Step 1.2.4.2

Simplify the right side.

Tap for more steps...

Step 1.2.4.2.1

Simplify each term.

Tap for more steps...

Step 1.2.4.2.1.1

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

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

Step 1.2.4.2.1.2

Multiply $4$ by $1$ .

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

Step 1.2.4.2.1.3

Divide $36$ by $4$ .

$e = 0 - 1 \cdot 9$

Step 1.2.4.2.1.4

Multiply $- 1$ by $9$ .

$e = 0 - 9$

Step 1.2.4.2.2

Subtract $9$ from $0$ .

$e = - 9$

Step 1.2.5

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

${(x - 3)}^{2} - 9$

Step 1.3

Substitute ${(x - 3)}^{2} - 9$ for $x^{2} - 6 x$ in the equation $x^{2} - y^{2} - 6 x - 4 y = 11$ .

${(x - 3)}^{2} - 9 - y^{2} - 4 y = 11$

Step 1.4

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

${(x - 3)}^{2} - y^{2} - 4 y = 11 + 9$

Step 1.5

Complete the square for $- y^{2} - 4 y$ .

Tap for more steps...

Step 1.5.1

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

$a = - 1$

$b = - 4$

$c = 0$

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 $d = \frac{b}{2 a}$ .

Tap for more steps...

Step 1.5.3.1

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

$d = \frac{- 4}{2 \cdot - 1}$

Step 1.5.3.2

Simplify the right side.

Tap for more steps...

Step 1.5.3.2.1

Cancel the common factor of $- 4$ and $2$ .

Tap for more steps...

Step 1.5.3.2.1.1

Factor $2$ out of $- 4$ .

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

Step 1.5.3.2.1.2

Move the negative one from the denominator of $\frac{- 2}{- 1}$ .

$d = - 1 \cdot - 2$

Step 1.5.3.2.2

Rewrite $- 1 \cdot - 2$ as $- - 2$ .

$d = - - 2$

Step 1.5.3.2.3

Multiply $- 1$ by $- 2$ .

$d = 2$

Step 1.5.4

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

Tap for more steps...

Step 1.5.4.1

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

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

Step 1.5.4.2

Simplify the right side.

Tap for more steps...

Step 1.5.4.2.1

Simplify each term.

Tap for more steps...

Step 1.5.4.2.1.1

Cancel the common factor of ${(- 4)}^{2}$ and $4$ .

Tap for more steps...

Step 1.5.4.2.1.1.1

Rewrite $- 4$ as $- 1 (4)$ .

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

Step 1.5.4.2.1.1.2

Apply the product rule to $- 1 (4)$ .

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

Step 1.5.4.2.1.1.3

Raise $- 1$ to the power of $2$ .

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

Step 1.5.4.2.1.1.4

Multiply $4^{2}$ by $1$ .

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

Step 1.5.4.2.1.1.5

Factor $4$ out of $4^{2}$ .

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

Step 1.5.4.2.1.1.6

Move the negative one from the denominator of $\frac{4}{- 1}$ .

$e = 0 - (- 1 \cdot 4)$

Step 1.5.4.2.1.2

Multiply $- (- 1 \cdot 4)$ .

Tap for more steps...

Step 1.5.4.2.1.2.1

Multiply $- 1$ by $4$ .

$e = 0 - - 4$

Step 1.5.4.2.1.2.2

Multiply $- 1$ by $- 4$ .

$e = 0 + 4$

Step 1.5.4.2.2

Add $0$ and $4$ .

$e = 4$

Step 1.5.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(y + 2)}^{2} + 4$ .

$- {(y + 2)}^{2} + 4$

Step 1.6

Substitute $- {(y + 2)}^{2} + 4$ for $- y^{2} - 4 y$ in the equation $x^{2} - y^{2} - 6 x - 4 y = 11$ .

${(x - 3)}^{2} - {(y + 2)}^{2} + 4 = 11 + 9$

Step 1.7

Move $4$ to the right side of the equation by adding $4$ to both sides.

${(x - 3)}^{2} - {(y + 2)}^{2} = 11 + 9 - 4$

Step 1.8

Simplify $11 + 9 - 4$ .

Tap for more steps...

Step 1.8.1

Add $11$ and $9$ .

${(x - 3)}^{2} - {(y + 2)}^{2} = 20 - 4$

Step 1.8.2

Subtract $4$ from $20$ .

${(x - 3)}^{2} - {(y + 2)}^{2} = 16$

Step 1.9

Divide each term by $16$ to make the right side equal to one.

$\frac{{(x - 3)}^{2}}{16} - \frac{{(y + 2)}^{2}}{16} = \frac{16}{16}$

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 $1$ .

$\frac{{(x - 3)}^{2}}{16} - \frac{{(y + 2)}^{2}}{16} = 1$

Step 2

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

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

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, $a$ .

$a = 4$

$b = 4$

$k = - 2$

$h = 3$

Step 4

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

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

Step 5

Simplify to find the first asymptote.

Tap for more steps...

Step 5.1

Remove parentheses.

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

Step 5.2

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

Tap for more steps...

Step 5.2.1

Simplify each term.

Tap for more steps...

Step 5.2.1.1

Multiply $x - (3)$ by $1$ .

$y = x - (3) - 2$

Step 5.2.1.2

Multiply $- 1$ by $3$ .

$y = x - 3 - 2$

Step 5.2.2

Subtract $2$ from $- 3$ .

$y = x - 5$

Step 6

Simplify to find the second asymptote.

Tap for more steps...

Step 6.1

Remove parentheses.

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

Step 6.2

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

Tap for more steps...

Step 6.2.1

Simplify each term.

Tap for more steps...

Step 6.2.1.1

Multiply $- 1$ by $3$ .

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

Step 6.2.1.2

Apply the distributive property.

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

Step 6.2.1.3

Rewrite $- 1 x$ as $- x$ .

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

Step 6.2.1.4

Multiply $- 1$ by $- 3$ .

$y = - x + 3 - 2$

Step 6.2.2

Subtract $2$ from $3$ .

$y = - x + 1$

Step 7

This hyperbola has two asymptotes.

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

Step 8

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

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

Step 9