Find the Asymptotes x^2-y^2-4x-12y-33=0

$x^{2} - y^{2} - 4 x - 12 y - 33 = 0$

Step 1

Find the standard form of the hyperbola.

Add $33$ to both sides of the equation.

$x^{2} - y^{2} - 4 x - 12 y = 33$

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

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

$a = 1$

$b = - 4$

$c = 0$

Consider the vertex form of a parabola.

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

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

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

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

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

Factor $2$ out of $- 4$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 2$ by $1$ .

$d = - 2$

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

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

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

Simplify the right side.

Simplify each term.

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

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

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

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

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

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

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

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

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

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

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

Cancel the common factors.

Factor $4$ out of $4 \cdot 1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $4$ by $1$ .

$e = 0 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = 0 - 4$

Subtract $4$ from $0$ .

$e = - 4$

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

${(x - 2)}^{2} - 4$

Substitute ${(x - 2)}^{2} - 4$ for $x^{2} - 4 x$ in the equation $x^{2} - y^{2} - 4 x - 12 y = 33$ .

${(x - 2)}^{2} - 4 - y^{2} - 12 y = 33$

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

${(x - 2)}^{2} - y^{2} - 12 y = 33 + 4$

Complete the square for $- y^{2} - 12 y$ .

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

$a = - 1$

$b = - 12$

$c = 0$

Consider the vertex form of a parabola.

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

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

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

$d = \frac{- 12}{2 \cdot - 1}$

Simplify the right side.

Cancel the common factor of $- 12$ and $2$ .

Factor $2$ out of $- 12$ .

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

Move the negative one from the denominator of $\frac{- 6}{- 1}$ .

$d = - 1 \cdot - 6$

Rewrite $- 1 \cdot - 6$ as $- - 6$ .

$d = - - 6$

Multiply $- 1$ by $- 6$ .

$d = 6$

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

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

$e = 0 - \frac{{(- 12)}^{2}}{4 \cdot - 1}$

Simplify the right side.

Simplify each term.

Raise $- 12$ to the power of $2$ .

$e = 0 - \frac{144}{4 \cdot - 1}$

Multiply $4$ by $- 1$ .

$e = 0 - \frac{144}{- 4}$

Divide $144$ by $- 4$ .

$e = 0 - - 36$

Multiply $- 1$ by $- 36$ .

$e = 0 + 36$

Add $0$ and $36$ .

$e = 36$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(y + 6)}^{2} + 36$ .

$- {(y + 6)}^{2} + 36$

Substitute $- {(y + 6)}^{2} + 36$ for $- y^{2} - 12 y$ in the equation $x^{2} - y^{2} - 4 x - 12 y = 33$ .

${(x - 2)}^{2} - {(y + 6)}^{2} + 36 = 33 + 4$

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

${(x - 2)}^{2} - {(y + 6)}^{2} = 33 + 4 - 36$

Simplify $33 + 4 - 36$ .

Add $33$ and $4$ .

${(x - 2)}^{2} - {(y + 6)}^{2} = 37 - 36$

Subtract $36$ from $37$ .

${(x - 2)}^{2} - {(y + 6)}^{2} = 1$

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, $a$ .

$a = 1$

$b = 1$

$k = - 6$

$h = 2$

Step 4

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

$y = \pm 1 \cdot (x - (2)) - 6$

Step 5

Simplify to find the first asymptote.

Remove parentheses.

$y = 1 \cdot (x - (2)) - 6$

Simplify $1 \cdot (x - (2)) - 6$ .

Simplify each term.

Multiply $x - (2)$ by $1$ .

$y = x - (2) - 6$

Multiply $- 1$ by $2$ .

$y = x - 2 - 6$

Subtract $6$ from $- 2$ .

$y = x - 8$

Step 6

Simplify to find the second asymptote.

Remove parentheses.

$y = - 1 \cdot (x - (2)) - 6$

Simplify $- 1 \cdot (x - (2)) - 6$ .

Simplify each term.

Multiply $- 1$ by $2$ .

$y = - 1 \cdot (x - 2) - 6$

Apply the distributive property.

$y = - 1 x - 1 \cdot - 2 - 6$

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

$y = - x - 1 \cdot - 2 - 6$

Multiply $- 1$ by $- 2$ .

$y = - x + 2 - 6$

Subtract $6$ from $2$ .

$y = - x - 4$

Step 7

This hyperbola has two asymptotes.

$y = x - 8, y = - x - 4$

Step 8

The asymptotes are $y = x - 8$ and $y = - x - 4$ .

Asymptotes: $y = x - 8, y = - x - 4$

Step 9

Find the Asymptotes x^2-y^2-4x-12y-33=0

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