Examples

4 x^{2} + y^{2} - 16 x + 2 y + 13 = 0

$4 x^{2} + y^{2} - 16 x + 2 y + 13 = 0$

Step 1

Subtract

13

$13$ from both sides of the equation.

4 x^{2} + y^{2} - 16 x + 2 y = - 13

$4 x^{2} + y^{2} - 16 x + 2 y = - 13$

Step 2

Complete the square for

4 x^{2} - 16 x

$4 x^{2} - 16 x$ .

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Step 2.1

Use the form

a x^{2} + b x + c

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

a

$a$ ,

b

$b$ , and

c

$c$ .

a = 4

$a = 4$

b = - 16

$b = - 16$

c = 0

$c = 0$

Step 2.2

Consider the vertex form of a parabola.

a {(x + d)}^{2} + e

$a {(x + d)}^{2} + e$

Step 2.3

Find the value of

d

$d$ using the formula

d = \frac{b}{2 a}

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

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Step 2.3.1

Substitute the values of

a

$a$ and

b

$b$ into the formula

d = \frac{b}{2 a}

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

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

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

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

Cancel the common factor of

- 16

$- 16$ and

2

$2$ .

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Step 2.3.2.1.1

Factor

2

$2$ out of

- 16

$- 16$ .

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

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

Step 2.3.2.1.2

Cancel the common factors.

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Step 2.3.2.1.2.1

Factor

2

$2$ out of

2 \cdot 4

$2 \cdot 4$ .

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

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

Step 2.3.2.1.2.2

Cancel the common factor.

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

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

Step 2.3.2.1.2.3

Rewrite the expression.

d = \frac{- 8}{4}

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

d = \frac{- 8}{4}

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

d = \frac{- 8}{4}

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

Step 2.3.2.2

Cancel the common factor of

- 8

$- 8$ and

4

$4$ .

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Step 2.3.2.2.1

Factor

4

$4$ out of

- 8

$- 8$ .

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

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

Step 2.3.2.2.2

Cancel the common factors.

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Step 2.3.2.2.2.1

Factor

4

$4$ out of

4

$4$ .

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

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

Step 2.3.2.2.2.2

Cancel the common factor.

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

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

Step 2.3.2.2.2.3

Rewrite the expression.

d = \frac{- 2}{1}

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

Step 2.3.2.2.2.4

Divide

- 2

$- 2$ by

1

$1$ .

d = - 2

$d = - 2$

d = - 2

$d = - 2$

d = - 2

$d = - 2$

d = - 2

$d = - 2$

d = - 2

$d = - 2$

Step 2.4

Find the value of

e

$e$ using the formula

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

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

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Step 2.4.1

Substitute the values of

c

$c$ ,

b

$b$ and

a

$a$ into the formula

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

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

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

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

Step 2.4.2

Simplify the right side.

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Step 2.4.2.1

Simplify each term.

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Step 2.4.2.1.1

Raise

- 16

$- 16$ to the power of

2

$2$ .

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

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

Step 2.4.2.1.2

Multiply

4

$4$ by

4

$4$ .

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

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

Step 2.4.2.1.3

Divide

256

$256$ by

16

$16$ .

e = 0 - 1 \cdot 16

$e = 0 - 1 \cdot 16$

Step 2.4.2.1.4

Multiply

- 1

$- 1$ by

16

$16$ .

e = 0 - 16

$e = 0 - 16$

e = 0 - 16

$e = 0 - 16$

Step 2.4.2.2

Subtract

16

$16$ from

0

$0$ .

e = - 16

$e = - 16$

e = - 16

$e = - 16$

e = - 16

$e = - 16$

Step 2.5

Substitute the values of

a

$a$ ,

d

$d$ , and

e

$e$ into the vertex form

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

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

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

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

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

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

Step 3

Substitute

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

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

4 x^{2} - 16 x

$4 x^{2} - 16 x$ in the equation

4 x^{2} + y^{2} - 16 x + 2 y = - 13

$4 x^{2} + y^{2} - 16 x + 2 y = - 13$ .

4 {(x - 2)}^{2} - 16 + y^{2} + 2 y = - 13

$4 {(x - 2)}^{2} - 16 + y^{2} + 2 y = - 13$

Step 4

Move

- 16

$- 16$ to the right side of the equation by adding

16

$16$ to both sides.

4 {(x - 2)}^{2} + y^{2} + 2 y = - 13 + 16

$4 {(x - 2)}^{2} + y^{2} + 2 y = - 13 + 16$

Step 5

Complete the square for

y^{2} + 2 y

$y^{2} + 2 y$ .

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Step 5.1

Use the form

a x^{2} + b x + c

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

a

$a$ ,

b

$b$ , and

c

$c$ .

a = 1

$a = 1$

b = 2

$b = 2$

c = 0

$c = 0$

Step 5.2

Consider the vertex form of a parabola.

a {(x + d)}^{2} + e

$a {(x + d)}^{2} + e$

Step 5.3

Find the value of

d

$d$ using the formula

d = \frac{b}{2 a}

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

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Step 5.3.1

Substitute the values of

a

$a$ and

b

$b$ into the formula

d = \frac{b}{2 a}

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

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

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

Step 5.3.2

Cancel the common factor of

2

$2$ .

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Step 5.3.2.1

Cancel the common factor.

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

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

Step 5.3.2.2

Rewrite the expression.

d = 1

$d = 1$

d = 1

$d = 1$

d = 1

$d = 1$

Step 5.4

Find the value of

e

$e$ using the formula

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

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

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Step 5.4.1

Substitute the values of

c

$c$ ,

b

$b$ and

a

$a$ into the formula

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

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

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

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

Step 5.4.2

Simplify the right side.

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Step 5.4.2.1

Simplify each term.

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Step 5.4.2.1.1

Raise

2

$2$ to the power of

2

$2$ .

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

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

Step 5.4.2.1.2

Multiply

4

$4$ by

1

$1$ .

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

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

Step 5.4.2.1.3

Cancel the common factor of

4

$4$ .

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Step 5.4.2.1.3.1

Cancel the common factor.

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

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

Step 5.4.2.1.3.2

Rewrite the expression.

e = 0 - 1 \cdot 1

$e = 0 - 1 \cdot 1$

e = 0 - 1 \cdot 1

$e = 0 - 1 \cdot 1$

Step 5.4.2.1.4

Multiply

- 1

$- 1$ by

1

$1$ .

e = 0 - 1

$e = 0 - 1$

e = 0 - 1

$e = 0 - 1$

Step 5.4.2.2

Subtract

1

$1$ from

0

$0$ .

e = - 1

$e = - 1$

e = - 1

$e = - 1$

e = - 1

$e = - 1$

Step 5.5

Substitute the values of

a

$a$ ,

d

$d$ , and

e

$e$ into the vertex form

{(y + 1)}^{2} - 1

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

{(y + 1)}^{2} - 1

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

{(y + 1)}^{2} - 1

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

Step 6

Substitute

{(y + 1)}^{2} - 1

${(y + 1)}^{2} - 1$ for

y^{2} + 2 y

$y^{2} + 2 y$ in the equation

4 x^{2} + y^{2} - 16 x + 2 y = - 13

$4 x^{2} + y^{2} - 16 x + 2 y = - 13$ .

4 {(x - 2)}^{2} + {(y + 1)}^{2} - 1 = - 13 + 16

$4 {(x - 2)}^{2} + {(y + 1)}^{2} - 1 = - 13 + 16$

Step 7

Move

- 1

$- 1$ to the right side of the equation by adding

1

$1$ to both sides.

4 {(x - 2)}^{2} + {(y + 1)}^{2} = - 13 + 16 + 1

$4 {(x - 2)}^{2} + {(y + 1)}^{2} = - 13 + 16 + 1$

Step 8

Simplify

- 13 + 16 + 1

$- 13 + 16 + 1$ .

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Step 8.1

Add

- 13

$- 13$ and

16

$16$ .

4 {(x - 2)}^{2} + {(y + 1)}^{2} = 3 + 1

$4 {(x - 2)}^{2} + {(y + 1)}^{2} = 3 + 1$

Step 8.2

Add

3

$3$ and

1

$1$ .

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

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

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

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

Step 9

Divide each term by

4

$4$ to make the right side equal to one.

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

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

Step 10

Simplify each term in the equation in order to set the right side equal to

1

$1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be

1

$1$ .

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

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

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