Find the Center -4x^2+y^2+16x+12y+16=0

$- 4 x^{2} + y^{2} + 16 x + 12 y + 16 = 0$

Step 1

Find the standard form of the hyperbola.

Subtract $16$ from both sides of the equation.

$- 4 x^{2} + y^{2} + 16 x + 12 y = - 16$

Complete the square for $- 4 x^{2} + 16 x$ .

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

$a = - 4$

$b = 16$

$c = 0$

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

$d = \frac{16}{2 \cdot - 4}$

Simplify the right side.

Cancel the common factor of $16$ and $2$ .

Factor $2$ out of $16$ .

$d = \frac{2 \cdot 8}{2 \cdot - 4}$

Cancel the common factors.

Factor $2$ out of $2 \cdot - 4$ .

$d = \frac{2 \cdot 8}{2 (- 4)}$

Cancel the common factor.

$d = \frac{2 \cdot 8}{2 \cdot - 4}$

Rewrite the expression.

$d = \frac{8}{- 4}$

Cancel the common factor of $8$ and $- 4$ .

Factor $4$ out of $8$ .

$d = \frac{4 (2)}{- 4}$

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

$d = - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$d = - 2$

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

$e = 0 - \frac{16^{2}}{4 \cdot - 4}$

Simplify the right side.

Simplify each term.

Raise $16$ to the power of $2$ .

$e = 0 - \frac{256}{4 \cdot - 4}$

Multiply $4$ by $- 4$ .

$e = 0 - \frac{256}{- 16}$

Divide $256$ by $- 16$ .

$e = 0 - - 16$

Multiply $- 1$ by $- 16$ .

$e = 0 + 16$

Add $0$ and $16$ .

$e = 16$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 4 {(x - 2)}^{2} + 16$ .

$- 4 {(x - 2)}^{2} + 16$

Substitute $- 4 {(x - 2)}^{2} + 16$ for $- 4 x^{2} + 16 x$ in the equation $- 4 x^{2} + y^{2} + 16 x + 12 y = - 16$ .

$- 4 {(x - 2)}^{2} + 16 + y^{2} + 12 y = - 16$

Move $16$ to the right side of the equation by adding $16$ to both sides.

$- 4 {(x - 2)}^{2} + y^{2} + 12 y = - 16 - 16$

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

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

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

Factor $2$ out of $12$ .

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

Cancel the common factors.

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

$d = \frac{2 \cdot 6}{2 (1)}$

Cancel the common factor.

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

Rewrite the expression.

$d = \frac{6}{1}$

Divide $6$ by $1$ .

$d = 6$

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 $4$ .

$e = 0 - 1 \cdot 36$

Multiply $- 1$ by $36$ .

$e = 0 - 36$

Subtract $36$ from $0$ .

$e = - 36$

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 $- 4 x^{2} + y^{2} + 16 x + 12 y = - 16$ .

$- 4 {(x - 2)}^{2} + {(y + 6)}^{2} - 36 = - 16 - 16$

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

$- 4 {(x - 2)}^{2} + {(y + 6)}^{2} = - 16 - 16 + 36$

Simplify $- 16 - 16 + 36$ .

Subtract $16$ from $- 16$ .

$- 4 {(x - 2)}^{2} + {(y + 6)}^{2} = - 32 + 36$

Add $- 32$ and $36$ .

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

Divide each term by $4$ to make the right side equal to one.

$- \frac{4 {(x - 2)}^{2}}{4} + \frac{{(y + 6)}^{2}}{4} = \frac{4}{4}$

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 $1$ .

$\frac{{(y + 6)}^{2}}{4} - \frac{{(x - 2)}^{2}}{1} = 1$

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.

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

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, $a$ .

$a = 2$

$b = 1$

$k = - 6$

$h = 2$

Step 4

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

$(2, - 6)$

Step 5

Find the Center -4x^2+y^2+16x+12y+16=0

Related Questions

Support LCMGCF.com by disabling your adblocker.