Find the Center x^2+8x+4y^2-40y+112=0

$x^{2} + 8 x + 4 y^{2} - 40 y + 112 = 0$

Step 1

Find the standard form of the ellipse.

Subtract $112$ from both sides of the equation.

$x^{2} + 8 x + 4 y^{2} - 40 y = - 112$

Complete the square for $x^{2} + 8 x$ .

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

$a = 1$

$b = 8$

$c = 0$

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

$d = \frac{8}{2 \cdot 1}$

Cancel the common factor of $8$ and $2$ .

Factor $2$ out of $8$ .

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

Cancel the common factors.

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

$d = \frac{2 \cdot 4}{2 (1)}$

Cancel the common factor.

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

Rewrite the expression.

$d = \frac{4}{1}$

Divide $4$ by $1$ .

$d = 4$

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

$e = 0 - \frac{8^{2}}{4 \cdot 1}$

Simplify the right side.

Simplify each term.

Raise $8$ to the power of $2$ .

$e = 0 - \frac{64}{4 \cdot 1}$

Multiply $4$ by $1$ .

$e = 0 - \frac{64}{4}$

Divide $64$ by $4$ .

$e = 0 - 1 \cdot 16$

Multiply $- 1$ by $16$ .

$e = 0 - 16$

Subtract $16$ from $0$ .

$e = - 16$

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

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

Substitute ${(x + 4)}^{2} - 16$ for $x^{2} + 8 x$ in the equation $x^{2} + 8 x + 4 y^{2} - 40 y = - 112$ .

${(x + 4)}^{2} - 16 + 4 y^{2} - 40 y = - 112$

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

${(x + 4)}^{2} + 4 y^{2} - 40 y = - 112 + 16$

Complete the square for $4 y^{2} - 40 y$ .

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

$a = 4$

$b = - 40$

$c = 0$

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

$d = \frac{- 40}{2 \cdot 4}$

Simplify the right side.

Cancel the common factor of $- 40$ and $2$ .

Factor $2$ out of $- 40$ .

$d = \frac{2 \cdot - 20}{2 \cdot 4}$

Cancel the common factors.

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

$d = \frac{2 \cdot - 20}{2 (4)}$

Cancel the common factor.

$d = \frac{2 \cdot - 20}{2 \cdot 4}$

Rewrite the expression.

$d = \frac{- 20}{4}$

Cancel the common factor of $- 20$ and $4$ .

Factor $4$ out of $- 20$ .

$d = \frac{4 \cdot - 5}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$d = \frac{4 \cdot - 5}{4 (1)}$

Cancel the common factor.

$d = \frac{4 \cdot - 5}{4 \cdot 1}$

Rewrite the expression.

$d = \frac{- 5}{1}$

Divide $- 5$ by $1$ .

$d = - 5$

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

$e = 0 - \frac{{(- 40)}^{2}}{4 \cdot 4}$

Simplify the right side.

Simplify each term.

Raise $- 40$ to the power of $2$ .

$e = 0 - \frac{1600}{4 \cdot 4}$

Multiply $4$ by $4$ .

$e = 0 - \frac{1600}{16}$

Divide $1600$ by $16$ .

$e = 0 - 1 \cdot 100$

Multiply $- 1$ by $100$ .

$e = 0 - 100$

Subtract $100$ from $0$ .

$e = - 100$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $4 {(y - 5)}^{2} - 100$ .

$4 {(y - 5)}^{2} - 100$

Substitute $4 {(y - 5)}^{2} - 100$ for $4 y^{2} - 40 y$ in the equation $x^{2} + 8 x + 4 y^{2} - 40 y = - 112$ .

${(x + 4)}^{2} + 4 {(y - 5)}^{2} - 100 = - 112 + 16$

Move $- 100$ to the right side of the equation by adding $100$ to both sides.

${(x + 4)}^{2} + 4 {(y - 5)}^{2} = - 112 + 16 + 100$

Simplify $- 112 + 16 + 100$ .

Add $- 112$ and $16$ .

${(x + 4)}^{2} + 4 {(y - 5)}^{2} = - 96 + 100$

Add $- 96$ and $100$ .

${(x + 4)}^{2} + 4 {(y - 5)}^{2} = 4$

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

$\frac{{(x + 4)}^{2}}{4} + \frac{4 {(y - 5)}^{2}}{4} = \frac{4}{4}$

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 $1$ .

$\frac{{(x + 4)}^{2}}{4} + {(y - 5)}^{2} = 1$

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.

$a = 2$

$b = 1$

$k = 5$

$h = - 4$

Step 4

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

$(- 4, 5)$

Step 5

Find the Center x^2+8x+4y^2-40y+112=0

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