Find the Center -4x^2+9y^2+16x-18y+29=0

$- 4 x^{2} + 9 y^{2} + 16 x - 18 y + 29 = 0$

Step 1

Find the standard form of the hyperbola.

Subtract $29$ from both sides of the equation.

$- 4 x^{2} + 9 y^{2} + 16 x - 18 y = - 29$

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 $- 4 x^{2} + 9 y^{2} + 16 x - 18 y = - 29$ .

$- 4 {(x - 2)}^{2} + 16 + 9 y^{2} - 18 y = - 29$

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

$- 4 {(x - 2)}^{2} + 9 y^{2} - 18 y = - 29 - 16$

Complete the square for $9 y^{2} - 18 y$ .

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

$a = 9$

$b = - 18$

$c = 0$

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

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

Simplify the right side.

Cancel the common factor of $- 18$ and $2$ .

Factor $2$ out of $- 18$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $9$ out of $- 9$ .

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

Cancel the common factors.

Factor $9$ out of $9$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- 1$ 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 $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 18)}^{2}}{4 \cdot 9}$

Simplify the right side.

Simplify each term.

Raise $- 18$ to the power of $2$ .

$e = 0 - \frac{324}{4 \cdot 9}$

Multiply $4$ by $9$ .

$e = 0 - \frac{324}{36}$

Divide $324$ by $36$ .

$e = 0 - 1 \cdot 9$

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 $9 {(y - 1)}^{2} - 9$ .

$9 {(y - 1)}^{2} - 9$

Substitute $9 {(y - 1)}^{2} - 9$ for $9 y^{2} - 18 y$ in the equation $- 4 x^{2} + 9 y^{2} + 16 x - 18 y = - 29$ .

$- 4 {(x - 2)}^{2} + 9 {(y - 1)}^{2} - 9 = - 29 - 16$

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

$- 4 {(x - 2)}^{2} + 9 {(y - 1)}^{2} = - 29 - 16 + 9$

Simplify $- 29 - 16 + 9$ .

Subtract $16$ from $- 29$ .

$- 4 {(x - 2)}^{2} + 9 {(y - 1)}^{2} = - 45 + 9$

Add $- 45$ and $9$ .

$- 4 {(x - 2)}^{2} + 9 {(y - 1)}^{2} = - 36$

Flip the sign on each term of the equation so the term on the right side is positive.

$4 {(x - 2)}^{2} - 9 {(y - 1)}^{2} = 36$

Divide each term by $36$ to make the right side equal to one.

$\frac{4 {(x - 2)}^{2}}{36} - \frac{9 {(y - 1)}^{2}}{36} = \frac{36}{36}$

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 $1$ .

$\frac{{(x - 2)}^{2}}{9} - \frac{{(y - 1)}^{2}}{4} = 1$

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, $a$ .

$a = 3$

$b = 2$

$k = 1$

$h = 2$

Step 4

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

$(2, 1)$

Step 5

Find the Center -4x^2+9y^2+16x-18y+29=0

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