Find the Center 4x^2+y^2-48x+4y+48=0

$4 x^{2} + y^{2} - 48 x + 4 y + 48 = 0$

Step 1

Find the standard form of the ellipse.

Subtract $48$ from both sides of the equation.

$4 x^{2} + y^{2} - 48 x + 4 y = - 48$

Complete the square for $4 x^{2} - 48 x$ .

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

$a = 4$

$b = - 48$

$c = 0$

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

$d = \frac{- 48}{2 \cdot 4}$

Simplify the right side.

Cancel the common factor of $- 48$ and $2$ .

Factor $2$ out of $- 48$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $- 24$ and $4$ .

Factor $4$ out of $- 24$ .

$d = \frac{4 \cdot - 6}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$d = \frac{4 \cdot - 6}{4 (1)}$

Cancel the common factor.

$d = \frac{4 \cdot - 6}{4 \cdot 1}$

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

$e = 0 - \frac{{(- 48)}^{2}}{4 \cdot 4}$

Simplify the right side.

Simplify each term.

Raise $- 48$ to the power of $2$ .

$e = 0 - \frac{2304}{4 \cdot 4}$

Multiply $4$ by $4$ .

$e = 0 - \frac{2304}{16}$

Divide $2304$ by $16$ .

$e = 0 - 1 \cdot 144$

Multiply $- 1$ by $144$ .

$e = 0 - 144$

Subtract $144$ from $0$ .

$e = - 144$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $4 {(x - 6)}^{2} - 144$ .

$4 {(x - 6)}^{2} - 144$

Substitute $4 {(x - 6)}^{2} - 144$ for $4 x^{2} - 48 x$ in the equation $4 x^{2} + y^{2} - 48 x + 4 y = - 48$ .

$4 {(x - 6)}^{2} - 144 + y^{2} + 4 y = - 48$

Move $- 144$ to the right side of the equation by adding $144$ to both sides.

$4 {(x - 6)}^{2} + y^{2} + 4 y = - 48 + 144$

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

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

$a = 1$

$b = 4$

$c = 0$

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

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

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factors.

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $2$ by $1$ .

$d = 2$

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

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $4^{2}$ and $4$ .

Factor $4$ out of $4^{2}$ .

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

Cancel the common factors.

Factor $4$ out of $4 \cdot 1$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $4$ by $1$ .

$e = 0 - 1 \cdot 4$

Multiply $- 1$ by $4$ .

$e = 0 - 4$

Subtract $4$ from $0$ .

$e = - 4$

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

${(y + 2)}^{2} - 4$

Substitute ${(y + 2)}^{2} - 4$ for $y^{2} + 4 y$ in the equation $4 x^{2} + y^{2} - 48 x + 4 y = - 48$ .

$4 {(x - 6)}^{2} + {(y + 2)}^{2} - 4 = - 48 + 144$

Move $- 4$ to the right side of the equation by adding $4$ to both sides.

$4 {(x - 6)}^{2} + {(y + 2)}^{2} = - 48 + 144 + 4$

Simplify $- 48 + 144 + 4$ .

Add $- 48$ and $144$ .

$4 {(x - 6)}^{2} + {(y + 2)}^{2} = 96 + 4$

Add $96$ and $4$ .

$4 {(x - 6)}^{2} + {(y + 2)}^{2} = 100$

Divide each term by $100$ to make the right side equal to one.

$\frac{4 {(x - 6)}^{2}}{100} + \frac{{(y + 2)}^{2}}{100} = \frac{100}{100}$

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 $1$ .

$\frac{{(x - 6)}^{2}}{25} + \frac{{(y + 2)}^{2}}{100} = 1$

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.

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

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.

$a = 10$

$b = 5$

$k = - 2$

$h = 6$

Step 4

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

$(6, - 2)$

Step 5

Find the Center 4x^2+y^2-48x+4y+48=0

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