Algebra Examples

$2 y^{2} + x + 20 y + 51 = 0$

Step 1

Rewrite the equation in vertex form.

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Step 1.1

Move all terms not containing $x$ to the right side of the equation.

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Step 1.1.1

Subtract $2 y^{2}$ from both sides of the equation.

$x + 20 y + 51 = - 2 y^{2}$

Step 1.1.2

Subtract $20 y$ from both sides of the equation.

$x + 51 = - 2 y^{2} - 20 y$

Step 1.1.3

Subtract $51$ from both sides of the equation.

$x = - 2 y^{2} - 20 y - 51$

Step 1.2

Complete the square for $- 2 y^{2} - 20 y - 51$ .

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Step 1.2.1

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

$a = - 2$

$b = - 20$

$c = - 51$

Step 1.2.2

Consider the vertex form of a parabola.

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

Step 1.2.3

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

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Step 1.2.3.1

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

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

Step 1.2.3.2

Simplify the right side.

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Step 1.2.3.2.1

Cancel the common factor of $- 20$ and $2$ .

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Step 1.2.3.2.1.1

Factor $2$ out of $- 20$ .

$d = \frac{2 \cdot - 10}{2 \cdot - 2}$

Step 1.2.3.2.1.2

Cancel the common factors.

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Step 1.2.3.2.1.2.1

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

$d = \frac{2 \cdot - 10}{2 (- 2)}$

Step 1.2.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot - 10}{2 \cdot - 2}$

Step 1.2.3.2.1.2.3

Rewrite the expression.

$d = \frac{- 10}{- 2}$

Step 1.2.3.2.2

Cancel the common factor of $- 10$ and $- 2$ .

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Step 1.2.3.2.2.1

Factor $- 2$ out of $- 10$ .

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

Step 1.2.3.2.2.2

Cancel the common factors.

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Step 1.2.3.2.2.2.1

Factor $- 2$ out of $- 2$ .

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

Step 1.2.3.2.2.2.2

Cancel the common factor.

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

Step 1.2.3.2.2.2.3

Rewrite the expression.

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

Step 1.2.3.2.2.2.4

Divide $5$ by $1$ .

$d = 5$

Step 1.2.4

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

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Step 1.2.4.1

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

$e = - 51 - \frac{{(- 20)}^{2}}{4 \cdot - 2}$

Step 1.2.4.2

Simplify the right side.

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Step 1.2.4.2.1

Simplify each term.

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Step 1.2.4.2.1.1

Raise $- 20$ to the power of $2$ .

$e = - 51 - \frac{400}{4 \cdot - 2}$

Step 1.2.4.2.1.2

Multiply $4$ by $- 2$ .

$e = - 51 - \frac{400}{- 8}$

Step 1.2.4.2.1.3

Divide $400$ by $- 8$ .

$e = - 51 - - 50$

Step 1.2.4.2.1.4

Multiply $- 1$ by $- 50$ .

$e = - 51 + 50$

Step 1.2.4.2.2

Add $- 51$ and $50$ .

$e = - 1$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- 2 {(y + 5)}^{2} - 1$ .

$- 2 {(y + 5)}^{2} - 1$

Step 1.3

Set $x$ equal to the new right side.

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

Step 2

Use the vertex form, $x = a {(y - k)}^{2} + h$ , to determine the values of $a$ , $h$ , and $k$ .

$a = - 2$

$h = - 1$

$k = - 5$

Step 3

Find the vertex $(h, k)$ .

$(- 1, - 5)$

Step 4

Find $p$ , the distance from the vertex to the focus.

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Step 4.1

Find the distance from the vertex to a focus of the parabola by using the following formula.

$\frac{1}{4 a}$

Step 4.2

Substitute the value of $a$ into the formula.

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

Step 4.3

Simplify.

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Step 4.3.1

Multiply $4$ by $- 2$ .

$\frac{1}{- 8}$

Step 4.3.2

Move the negative in front of the fraction.

$- \frac{1}{8}$

Step 5

Find the directrix.

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Step 5.1

The directrix of a parabola is the vertical line found by subtracting $p$ from the x-coordinate $h$ of the vertex if the parabola opens left or right.

$x = h - p$

Step 5.2

Substitute the known values of $p$ and $h$ into the formula and simplify.

$x = - \frac{7}{8}$

Step 6