Find the Center x^2+6x+4y^2-24y+9=0

$x^{2} + 6 x + 4 y^{2} - 24 y + 9 = 0$

Step 1

Find the standard form of the ellipse.

Subtract $9$ from both sides of the equation.

$x^{2} + 6 x + 4 y^{2} - 24 y = - 9$

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 $x^{2} + 6 x + 4 y^{2} - 24 y = - 9$ .

${(x + 3)}^{2} - 9 + 4 y^{2} - 24 y = - 9$

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

${(x + 3)}^{2} + 4 y^{2} - 24 y = - 9 + 9$

Complete the square for $4 y^{2} - 24 y$ .

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

$a = 4$

$b = - 24$

$c = 0$

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

$d = \frac{- 24}{2 \cdot 4}$

Simplify the right side.

Cancel the common factor of $- 24$ and $2$ .

Factor $2$ out of $- 24$ .

$d = \frac{2 \cdot - 12}{2 \cdot 4}$

Cancel the common factors.

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

$d = \frac{2 \cdot - 12}{2 (4)}$

Cancel the common factor.

$d = \frac{2 \cdot - 12}{2 \cdot 4}$

Rewrite the expression.

$d = \frac{- 12}{4}$

Cancel the common factor of $- 12$ and $4$ .

Factor $4$ out of $- 12$ .

$d = \frac{4 \cdot - 3}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$d = \frac{4 \cdot - 3}{4 (1)}$

Cancel the common factor.

$d = \frac{4 \cdot - 3}{4 \cdot 1}$

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

$e = 0 - \frac{{(- 24)}^{2}}{4 \cdot 4}$

Simplify the right side.

Simplify each term.

Raise $- 24$ to the power of $2$ .

$e = 0 - \frac{576}{4 \cdot 4}$

Multiply $4$ by $4$ .

$e = 0 - \frac{576}{16}$

Divide $576$ by $16$ .

$e = 0 - 1 \cdot 36$

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 $4 {(y - 3)}^{2} - 36$ .

$4 {(y - 3)}^{2} - 36$

Substitute $4 {(y - 3)}^{2} - 36$ for $4 y^{2} - 24 y$ in the equation $x^{2} + 6 x + 4 y^{2} - 24 y = - 9$ .

${(x + 3)}^{2} + 4 {(y - 3)}^{2} - 36 = - 9 + 9$

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

${(x + 3)}^{2} + 4 {(y - 3)}^{2} = - 9 + 9 + 36$

Simplify $- 9 + 9 + 36$ .

Add $- 9$ and $9$ .

${(x + 3)}^{2} + 4 {(y - 3)}^{2} = 0 + 36$

Add $0$ and $36$ .

${(x + 3)}^{2} + 4 {(y - 3)}^{2} = 36$

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

$\frac{{(x + 3)}^{2}}{36} + \frac{4 {(y - 3)}^{2}}{36} = \frac{36}{36}$

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 $1$ .

$\frac{{(x + 3)}^{2}}{36} + \frac{{(y - 3)}^{2}}{9} = 1$

Step 2

This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.

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

Step 3

Match the values in this ellipse to those of the standard form. The variable $a$ represents the radius of the major axis of the ellipse, $b$ represents the radius of the minor axis of the ellipse, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from the origin.

$a = 6$

$b = 3$

$k = 3$

$h = - 3$

Step 4

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

$(- 3, 3)$

Step 5

Find the Center x^2+6x+4y^2-24y+9=0

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