Find the Directrix y^2-12x+8y=-40

$y^{2} - 12 x + 8 y = - 40$

Step 1

Rewrite the equation in vertex form.

Isolate $x$ to the left side of the equation.

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

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

$- 12 x + 8 y = - 40 - y^{2}$

Subtract $8 y$ from both sides of the equation.

$- 12 x = - 40 - y^{2} - 8 y$

Divide each term in $- 12 x = - 40 - y^{2} - 8 y$ by $- 12$ and simplify.

Divide each term in $- 12 x = - 40 - y^{2} - 8 y$ by $- 12$ .

$\frac{- 12 x}{- 12} = \frac{- 40}{- 12} + \frac{- y^{2}}{- 12} + \frac{- 8 y}{- 12}$

Simplify the left side.

Cancel the common factor of $- 12$ .

Cancel the common factor.

$\frac{- 12 x}{- 12} = \frac{- 40}{- 12} + \frac{- y^{2}}{- 12} + \frac{- 8 y}{- 12}$

Divide $x$ by $1$ .

$x = \frac{- 40}{- 12} + \frac{- y^{2}}{- 12} + \frac{- 8 y}{- 12}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $- 40$ and $- 12$ .

Factor $- 4$ out of $- 40$ .

$x = \frac{- 4 (10)}{- 12} + \frac{- y^{2}}{- 12} + \frac{- 8 y}{- 12}$

Cancel the common factors.

Factor $- 4$ out of $- 12$ .

$x = \frac{- 4 \cdot 10}{- 4 \cdot 3} + \frac{- y^{2}}{- 12} + \frac{- 8 y}{- 12}$

Cancel the common factor.

$x = \frac{- 4 \cdot 10}{- 4 \cdot 3} + \frac{- y^{2}}{- 12} + \frac{- 8 y}{- 12}$

Rewrite the expression.

$x = \frac{10}{3} + \frac{- y^{2}}{- 12} + \frac{- 8 y}{- 12}$

Dividing two negative values results in a positive value.

$x = \frac{10}{3} + \frac{y^{2}}{12} + \frac{- 8 y}{- 12}$

Cancel the common factor of $- 8$ and $- 12$ .

Factor $- 4$ out of $- 8 y$ .

$x = \frac{10}{3} + \frac{y^{2}}{12} + \frac{- 4 (2 y)}{- 12}$

Cancel the common factors.

Factor $- 4$ out of $- 12$ .

$x = \frac{10}{3} + \frac{y^{2}}{12} + \frac{- 4 (2 y)}{- 4 (3)}$

Cancel the common factor.

$x = \frac{10}{3} + \frac{y^{2}}{12} + \frac{- 4 (2 y)}{- 4 \cdot 3}$

Rewrite the expression.

$x = \frac{10}{3} + \frac{y^{2}}{12} + \frac{2 y}{3}$

Complete the square for $\frac{10}{3} + \frac{y^{2}}{12} + \frac{2 y}{3}$ .

Move $\frac{10}{3}$ .

$\frac{y^{2}}{12} + \frac{2 y}{3} + \frac{10}{3}$

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

$a = \frac{1}{12}$

$b = \frac{2}{3}$

$c = \frac{10}{3}$

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

$d = \frac{\frac{2}{3}}{2 (\frac{1}{12})}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$d = \frac{2}{3} \cdot \frac{1}{2 (\frac{1}{12})}$

Cancel the common factor of $2$ .

Cancel the common factor.

$d = \frac{2}{3} \cdot \frac{1}{2 (\frac{1}{12})}$

Rewrite the expression.

$d = \frac{1}{3} \cdot \frac{1}{\frac{1}{12}}$

Multiply $\frac{1}{3}$ by $\frac{1}{\frac{1}{12}}$ .

$d = \frac{1}{3 (\frac{1}{12})}$

Combine $3$ and $\frac{1}{12}$ .

$d = \frac{1}{\frac{3}{12}}$

Cancel the common factor of $3$ and $12$ .

Factor $3$ out of $3$ .

$d = \frac{1}{\frac{3 (1)}{12}}$

Cancel the common factors.

Factor $3$ out of $12$ .

$d = \frac{1}{\frac{3 \cdot 1}{3 \cdot 4}}$

Cancel the common factor.

$d = \frac{1}{\frac{3 \cdot 1}{3 \cdot 4}}$

Rewrite the expression.

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

Multiply the numerator by the reciprocal of the denominator.

$d = 1 \cdot 4$

Multiply $4$ by $1$ .

$d = 4$

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

$e = \frac{10}{3} - \frac{{(\frac{2}{3})}^{2}}{4 (\frac{1}{12})}$

Simplify the right side.

Simplify each term.

Simplify the numerator.

Apply the product rule to $\frac{2}{3}$ .

$e = \frac{10}{3} - \frac{\frac{2^{2}}{3^{2}}}{4 (\frac{1}{12})}$

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

$e = \frac{10}{3} - \frac{\frac{4}{3^{2}}}{4 (\frac{1}{12})}$

Raise $3$ to the power of $2$ .

$e = \frac{10}{3} - \frac{\frac{4}{9}}{4 (\frac{1}{12})}$

Combine $4$ and $\frac{1}{12}$ .

$e = \frac{10}{3} - \frac{\frac{4}{9}}{\frac{4}{12}}$

Cancel the common factor of $4$ and $12$ .

Factor $4$ out of $4$ .

$e = \frac{10}{3} - \frac{\frac{4}{9}}{\frac{4 (1)}{12}}$

Cancel the common factors.

Factor $4$ out of $12$ .

$e = \frac{10}{3} - \frac{\frac{4}{9}}{\frac{4 \cdot 1}{4 \cdot 3}}$

Cancel the common factor.

$e = \frac{10}{3} - \frac{\frac{4}{9}}{\frac{4 \cdot 1}{4 \cdot 3}}$

Rewrite the expression.

$e = \frac{10}{3} - \frac{\frac{4}{9}}{\frac{1}{3}}$

Multiply the numerator by the reciprocal of the denominator.

$e = \frac{10}{3} - (\frac{4}{9} \cdot 3)$

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

$e = \frac{10}{3} - (\frac{4}{3 (3)} \cdot 3)$

Cancel the common factor.

$e = \frac{10}{3} - (\frac{4}{3 \cdot 3} \cdot 3)$

Rewrite the expression.

$e = \frac{10}{3} - \frac{4}{3}$

Combine the numerators over the common denominator.

$e = \frac{10 - 4}{3}$

Subtract $4$ from $10$ .

$e = \frac{6}{3}$

Divide $6$ by $3$ .

$e = 2$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $\frac{1}{12} {(y + 4)}^{2} + 2$ .

$\frac{1}{12} {(y + 4)}^{2} + 2$

Set $x$ equal to the new right side.

$x = \frac{1}{12} \cdot {(y + 4)}^{2} + 2$

Step 2

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

$a = \frac{1}{12}$

$h = 2$

$k = - 4$

Step 3

Find the vertex $(h, k)$ .

$(2, - 4)$

Step 4

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

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

$\frac{1}{4 a}$

Substitute the value of $a$ into the formula.

$\frac{1}{4 \cdot \frac{1}{12}}$

Simplify.

Combine $4$ and $\frac{1}{12}$ .

$\frac{1}{\frac{4}{12}}$

Cancel the common factor of $4$ and $12$ .

Factor $4$ out of $4$ .

$\frac{1}{\frac{4 (1)}{12}}$

Cancel the common factors.

Factor $4$ out of $12$ .

$\frac{1}{\frac{4 \cdot 1}{4 \cdot 3}}$

Cancel the common factor.

$\frac{1}{\frac{4 \cdot 1}{4 \cdot 3}}$

Rewrite the expression.

$\frac{1}{\frac{1}{3}}$

Multiply the numerator by the reciprocal of the denominator.

$1 \cdot 3$

Multiply $3$ by $1$ .

$3$

Step 5

Find the directrix.

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.

$x = h - p$

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

$x = - 1$

Step 6

Find the Directrix y^2-12x+8y=-40

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