Precalculus Examples

$2 x^{2} + 2 y^{2} - 10 x - 18 y - 1 = 0$

Step 1

Add $1$ to both sides of the equation.

$2 x^{2} + 2 y^{2} - 10 x - 18 y = 1$

Step 2

Divide both sides of the equation by $2$ .

$x^{2} + y^{2} - 5 x - 9 y = \frac{1}{2}$

Step 3

Complete the square for $x^{2} - 5 x$ .

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Step 3.1

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

$a = 1$

$b = - 5$

$c = 0$

Step 3.2

Consider the vertex form of a parabola.

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

Step 3.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 3.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 5}{2 \cdot 1}$

Step 3.3.2

Simplify the right side.

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Step 3.3.2.1

Multiply $2$ by $1$ .

$d = \frac{- 5}{2}$

Step 3.3.2.2

Move the negative in front of the fraction.

$d = - \frac{5}{2}$

Step 3.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 3.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 5)}^{2}}{4 \cdot 1}$

Step 3.4.2

Simplify the right side.

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Step 3.4.2.1

Simplify each term.

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Step 3.4.2.1.1

Raise $- 5$ to the power of $2$ .

$e = 0 - \frac{25}{4 \cdot 1}$

Step 3.4.2.1.2

Multiply $4$ by $1$ .

$e = 0 - \frac{25}{4}$

Step 3.4.2.2

Subtract $\frac{25}{4}$ from $0$ .

$e = - \frac{25}{4}$

Step 3.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x - \frac{5}{2})}^{2} - \frac{25}{4}$ .

${(x - \frac{5}{2})}^{2} - \frac{25}{4}$

Step 4

Substitute ${(x - \frac{5}{2})}^{2} - \frac{25}{4}$ for $x^{2} - 5 x$ in the equation $x^{2} + y^{2} - 5 x - 9 y = \frac{1}{2}$ .

${(x - \frac{5}{2})}^{2} - \frac{25}{4} + y^{2} - 9 y = \frac{1}{2}$

Step 5

Move $- \frac{25}{4}$ to the right side of the equation by adding $\frac{25}{4}$ to both sides.

${(x - \frac{5}{2})}^{2} + y^{2} - 9 y = \frac{1}{2} + \frac{25}{4}$

Step 6

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

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Step 6.1

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

$a = 1$

$b = - 9$

$c = 0$

Step 6.2

Consider the vertex form of a parabola.

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

Step 6.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 6.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{- 9}{2 \cdot 1}$

Step 6.3.2

Simplify the right side.

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Step 6.3.2.1

Multiply $2$ by $1$ .

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

Step 6.3.2.2

Move the negative in front of the fraction.

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

Step 6.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 6.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{{(- 9)}^{2}}{4 \cdot 1}$

Step 6.4.2

Simplify the right side.

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Step 6.4.2.1

Simplify each term.

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Step 6.4.2.1.1

Raise $- 9$ to the power of $2$ .

$e = 0 - \frac{81}{4 \cdot 1}$

Step 6.4.2.1.2

Multiply $4$ by $1$ .

$e = 0 - \frac{81}{4}$

Step 6.4.2.2

Subtract $\frac{81}{4}$ from $0$ .

$e = - \frac{81}{4}$

Step 6.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(y - \frac{9}{2})}^{2} - \frac{81}{4}$ .

${(y - \frac{9}{2})}^{2} - \frac{81}{4}$

Step 7

Substitute ${(y - \frac{9}{2})}^{2} - \frac{81}{4}$ for $y^{2} - 9 y$ in the equation $x^{2} + y^{2} - 5 x - 9 y = \frac{1}{2}$ .

${(x - \frac{5}{2})}^{2} + {(y - \frac{9}{2})}^{2} - \frac{81}{4} = \frac{1}{2} + \frac{25}{4}$

Step 8

Move $- \frac{81}{4}$ to the right side of the equation by adding $\frac{81}{4}$ to both sides.

${(x - \frac{5}{2})}^{2} + {(y - \frac{9}{2})}^{2} = \frac{1}{2} + \frac{25}{4} + \frac{81}{4}$

Step 9

Simplify $\frac{1}{2} + \frac{25}{4} + \frac{81}{4}$ .

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Step 9.1

Combine the numerators over the common denominator.

${(x - \frac{5}{2})}^{2} + {(y - \frac{9}{2})}^{2} = \frac{1}{2} + \frac{25 + 81}{4}$

Step 9.2

Add $25$ and $81$ .

${(x - \frac{5}{2})}^{2} + {(y - \frac{9}{2})}^{2} = \frac{1}{2} + \frac{106}{4}$

Step 9.3

Cancel the common factor of $106$ and $4$ .

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Step 9.3.1

Factor $2$ out of $106$ .

${(x - \frac{5}{2})}^{2} + {(y - \frac{9}{2})}^{2} = \frac{1}{2} + \frac{2 (53)}{4}$

Step 9.3.2

Cancel the common factors.

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Step 9.3.2.1

Factor $2$ out of $4$ .

${(x - \frac{5}{2})}^{2} + {(y - \frac{9}{2})}^{2} = \frac{1}{2} + \frac{2 \cdot 53}{2 \cdot 2}$

Step 9.3.2.2

Cancel the common factor.

${(x - \frac{5}{2})}^{2} + {(y - \frac{9}{2})}^{2} = \frac{1}{2} + \frac{2 \cdot 53}{2 \cdot 2}$

Step 9.3.2.3

Rewrite the expression.

${(x - \frac{5}{2})}^{2} + {(y - \frac{9}{2})}^{2} = \frac{1}{2} + \frac{53}{2}$

Step 9.4

Combine the numerators over the common denominator.

${(x - \frac{5}{2})}^{2} + {(y - \frac{9}{2})}^{2} = \frac{1 + 53}{2}$

Step 9.5

Simplify the expression.

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Step 9.5.1

Add $1$ and $53$ .

${(x - \frac{5}{2})}^{2} + {(y - \frac{9}{2})}^{2} = \frac{54}{2}$

Step 9.5.2

Divide $54$ by $2$ .

${(x - \frac{5}{2})}^{2} + {(y - \frac{9}{2})}^{2} = 27$