Find the Center x^2-y^2-6x-2y-1=0

$x^{2} - y^{2} - 6 x - 2 y - 1 = 0$

Find the standard form of the hyperbola.

Add $1$ to both sides of the equation.

$x^{2} - y^{2} - 6 x - 2 y = 1$

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

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

$a = 1$

$b = - 6$

$c = 0$

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

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

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

Factor $2$ out of $- 6$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 3$ by $1$ .

$d = - 3$

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 $e = c - \frac{b^{2}}{4 a}$ .

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $1$ .

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

Divide $36$ by $4$ .

$e = 0 - 1 \cdot 9$

Multiply $- 1$ by $9$ .

$e = 0 - 9$

Subtract $9$ from $0$ .

$e = - 9$

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

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

Substitute ${(x - 3)}^{2} - 9$ for $x^{2} - 6 x$ in the equation $x^{2} - y^{2} - 6 x - 2 y = 1$ .

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

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

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

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

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

$a = - 1$

$b = - 2$

$c = 0$

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

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

Simplify the right side.

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

Factor $2$ out of $- 2$ .

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

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

$d = - 1 \cdot - 1$

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

$d = - - 1$

Multiply $- 1$ by $- 1$ .

$d = 1$

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 $e = c - \frac{b^{2}}{4 a}$ .

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

Simplify the right side.

Simplify each term.

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

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

Multiply $4$ by $- 1$ .

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

Divide $4$ by $- 4$ .

$e = 0 - - 1$

Multiply $- 1$ by $- 1$ .

$e = 0 + 1$

Add $0$ and $1$ .

$e = 1$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- {(y + 1)}^{2} + 1$ .

$- {(y + 1)}^{2} + 1$

Substitute $- {(y + 1)}^{2} + 1$ for $- y^{2} - 2 y$ in the equation $x^{2} - y^{2} - 6 x - 2 y = 1$ .

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

Move $1$ to the right side of the equation by adding $1$ to both sides.

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

Simplify $1 + 9 - 1$ .

Add $1$ and $9$ .

${(x - 3)}^{2} - {(y + 1)}^{2} = 10 - 1$

Subtract $1$ from $10$ .

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

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

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

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 $1$ .

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

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

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

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, $a$ .

$a = 3$

$b = 3$

$k = - 1$

$h = 3$

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

$(3, - 1)$

Find the Center x^2-y^2-6x-2y-1=0

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