Precalculus Examples

$x^{2} + 5 y^{2} + x - 35 y + 56.5 = 0$

Step 1

Subtract $56.5$ from both sides of the equation.

$x^{2} + 5 y^{2} + x - 35 y = - 56.5$

Step 2

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

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Step 2.1

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

$a = 1$

$b = 1$

$c = 0$

Step 2.2

Consider the vertex form of a parabola.

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

Cancel the common factor of $1$ .

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Step 2.3.2.1

Cancel the common factor.

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

Step 2.3.2.2

Rewrite the expression.

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

Step 2.4

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

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Step 2.4.1

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

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

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

One to any power is one.

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

Step 2.4.2.1.2

Multiply $4$ by $1$ .

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

Step 2.4.2.2

Subtract $\frac{1}{4}$ from $0$ .

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

Step 2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + \frac{1}{2})}^{2} - \frac{1}{4}$ .

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

Step 3

Substitute ${(x + \frac{1}{2})}^{2} - \frac{1}{4}$ for $x^{2} + x$ in the equation $x^{2} + 5 y^{2} + x - 35 y = - 56.5$ .

${(x + \frac{1}{2})}^{2} - \frac{1}{4} + 5 y^{2} - 35 y = - 56.5$

Step 4

Move $- \frac{1}{4}$ to the right side of the equation by adding $\frac{1}{4}$ to both sides.

${(x + \frac{1}{2})}^{2} + 5 y^{2} - 35 y = - 56.5 + \frac{1}{4}$

Step 5

Complete the square for $5 y^{2} - 35 y$ .

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Step 5.1

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

$a = 5$

$b = - 35$

$c = 0$

Step 5.2

Consider the vertex form of a parabola.

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

Step 5.3

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

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Step 5.3.1

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

$d = \frac{- 35}{2 \cdot 5}$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Cancel the common factor of $- 35$ and $5$ .

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Step 5.3.2.1.1

Factor $5$ out of $- 35$ .

$d = \frac{5 \cdot - 7}{2 \cdot 5}$

Step 5.3.2.1.2

Cancel the common factors.

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Step 5.3.2.1.2.1

Factor $5$ out of $2 \cdot 5$ .

$d = \frac{5 \cdot - 7}{5 \cdot 2}$

Step 5.3.2.1.2.2

Cancel the common factor.

$d = \frac{5 \cdot - 7}{5 \cdot 2}$

Step 5.3.2.1.2.3

Rewrite the expression.

$d = \frac{- 7}{2}$

Step 5.3.2.2

Move the negative in front of the fraction.

$d = - \frac{7}{2}$

Step 5.4

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

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Step 5.4.1

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

$e = 0 - \frac{{(- 35)}^{2}}{4 \cdot 5}$

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 $- 35$ to the power of $2$ .

$e = 0 - \frac{1225}{4 \cdot 5}$

Step 5.4.2.1.2

Multiply $4$ by $5$ .

$e = 0 - \frac{1225}{20}$

Step 5.4.2.1.3

Cancel the common factor of $1225$ and $20$ .

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Step 5.4.2.1.3.1

Factor $5$ out of $1225$ .

$e = 0 - \frac{5 (245)}{20}$

Step 5.4.2.1.3.2

Cancel the common factors.

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Step 5.4.2.1.3.2.1

Factor $5$ out of $20$ .

$e = 0 - \frac{5 \cdot 245}{5 \cdot 4}$

Step 5.4.2.1.3.2.2

Cancel the common factor.

$e = 0 - \frac{5 \cdot 245}{5 \cdot 4}$

Step 5.4.2.1.3.2.3

Rewrite the expression.

$e = 0 - \frac{245}{4}$

Step 5.4.2.2

Subtract $\frac{245}{4}$ from $0$ .

$e = - \frac{245}{4}$

Step 5.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $5 {(y - \frac{7}{2})}^{2} - \frac{245}{4}$ .

$5 {(y - \frac{7}{2})}^{2} - \frac{245}{4}$

Step 6

Substitute $5 {(y - \frac{7}{2})}^{2} - \frac{245}{4}$ for $5 y^{2} - 35 y$ in the equation $x^{2} + 5 y^{2} + x - 35 y = - 56.5$ .

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} - \frac{245}{4} = - 56.5 + \frac{1}{4}$

Step 7

Move $- \frac{245}{4}$ to the right side of the equation by adding $\frac{245}{4}$ to both sides.

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = - 56.5 + \frac{1}{4} + \frac{245}{4}$

Step 8

Simplify $- 56.5 + \frac{1}{4} + \frac{245}{4}$ .

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Step 8.1

Combine the numerators over the common denominator.

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = - 56.5 + \frac{1 + 245}{4}$

Step 8.2

Add $1$ and $245$ .

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = - 56.5 + \frac{246}{4}$

Step 8.3

Cancel the common factor of $246$ and $4$ .

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Step 8.3.1

Factor $2$ out of $246$ .

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = - 56.5 + \frac{2 (123)}{4}$

Step 8.3.2

Cancel the common factors.

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Step 8.3.2.1

Factor $2$ out of $4$ .

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = - 56.5 + \frac{2 \cdot 123}{2 \cdot 2}$

Step 8.3.2.2

Cancel the common factor.

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = - 56.5 + \frac{2 \cdot 123}{2 \cdot 2}$

Step 8.3.2.3

Rewrite the expression.

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = - 56.5 + \frac{123}{2}$

Step 8.4

To write $- 56.5$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = - 56.5 \cdot \frac{2}{2} + \frac{123}{2}$

Step 8.5

Combine $- 56.5$ and $\frac{2}{2}$ .

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = \frac{- 56.5 \cdot 2}{2} + \frac{123}{2}$

Step 8.6

Combine the numerators over the common denominator.

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = \frac{- 56.5 \cdot 2 + 123}{2}$

Step 8.7

Simplify the numerator.

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Step 8.7.1

Multiply $- 56.5$ by $2$ .

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = \frac{- 113 + 123}{2}$

Step 8.7.2

Add $- 113$ and $123$ .

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = \frac{10}{2}$

Step 8.8

Divide $10$ by $2$ .

${(x + \frac{1}{2})}^{2} + 5 {(y - \frac{7}{2})}^{2} = 5$

Step 9

Divide each term by $5$ to make the right side equal to one.

$\frac{{(x + \frac{1}{2})}^{2}}{5} + \frac{5 {(y - \frac{7}{2})}^{2}}{5} = \frac{5}{5}$

Step 10

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 $1$ .

$\frac{{(x + \frac{1}{2})}^{2}}{\frac{1}{0.2}} + {(y - \frac{7}{2})}^{2} = 1$