Find the Directrix y^2-4y+12x+4=0

$y^{2} - 4 y + 12 x + 4 = 0$

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.

$- 4 y + 12 x + 4 = - y^{2}$

Add $4 y$ to both sides of the equation.

$12 x + 4 = - y^{2} + 4 y$

Subtract $4$ from both sides of the equation.

$12 x = - y^{2} + 4 y - 4$

Divide each term in $12 x = - y^{2} + 4 y - 4$ by $12$ and simplify.

Divide each term in $12 x = - y^{2} + 4 y - 4$ by $12$ .

$\frac{12 x}{12} = \frac{- y^{2}}{12} + \frac{4 y}{12} + \frac{- 4}{12}$

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

$\frac{12 x}{12} = \frac{- y^{2}}{12} + \frac{4 y}{12} + \frac{- 4}{12}$

Divide $x$ by $1$ .

$x = \frac{- y^{2}}{12} + \frac{4 y}{12} + \frac{- 4}{12}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$x = - \frac{y^{2}}{12} + \frac{4 y}{12} + \frac{- 4}{12}$

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

Factor $4$ out of $4 y$ .

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

Cancel the common factors.

Factor $4$ out of $12$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Factor $4$ out of $- 4$ .

$x = - \frac{y^{2}}{12} + \frac{y}{3} + \frac{4 (- 1)}{12}$

Cancel the common factors.

Factor $4$ out of $12$ .

$x = - \frac{y^{2}}{12} + \frac{y}{3} + \frac{4 \cdot - 1}{4 \cdot 3}$

Cancel the common factor.

$x = - \frac{y^{2}}{12} + \frac{y}{3} + \frac{4 \cdot - 1}{4 \cdot 3}$

Rewrite the expression.

$x = - \frac{y^{2}}{12} + \frac{y}{3} + \frac{- 1}{3}$

Move the negative in front of the fraction.

$x = - \frac{y^{2}}{12} + \frac{y}{3} - \frac{1}{3}$

Complete the square for $- \frac{y^{2}}{12} + \frac{y}{3} - \frac{1}{3}$ .

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

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

$b = \frac{1}{3}$

$c = - \frac{1}{3}$

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

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

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

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

Cancel the common factor of $1$ and $- 1$ .

Rewrite $1$ as $- 1 (- 1)$ .

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

Move the negative in front of the fraction.

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

Combine $2$ and $\frac{1}{12}$ .

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

Cancel the common factor of $2$ and $12$ .

Factor $2$ out of $2$ .

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

Cancel the common factors.

Factor $2$ out of $12$ .

$d = \frac{1}{3} (- \frac{1}{\frac{2 \cdot 1}{2 \cdot 6}})$

Cancel the common factor.

$d = \frac{1}{3} (- \frac{1}{\frac{2 \cdot 1}{2 \cdot 6}})$

Rewrite the expression.

$d = \frac{1}{3} (- \frac{1}{\frac{1}{6}})$

Multiply the numerator by the reciprocal of the denominator.

$d = \frac{1}{3} (- (1 \cdot 6))$

Multiply $- (1 \cdot 6)$ .

Multiply $6$ by $1$ .

$d = \frac{1}{3} (- 1 \cdot 6)$

Multiply $- 1$ by $6$ .

$d = \frac{1}{3} \cdot - 6$

Cancel the common factor of $3$ .

Factor $3$ out of $- 6$ .

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

Cancel the common factor.

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

Rewrite the expression.

$d = - 2$

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

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

Simplify the right side.

Simplify each term.

Simplify the numerator.

Apply the product rule to $\frac{1}{3}$ .

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

One to any power is one.

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

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

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

Simplify the denominator.

Multiply $- 1$ by $4$ .

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

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

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

Reduce the expression by cancelling the common factors.

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

Factor $4$ out of $- 4$ .

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

Cancel the common factors.

Factor $4$ out of $12$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Multiply the numerator by the reciprocal of the denominator.

$e = - \frac{1}{3} - (\frac{1}{9} (- 1 \cdot 3))$

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

$e = - \frac{1}{3} - (\frac{1}{3 (3)} (- 1 \cdot 3))$

Factor $3$ out of $- 1 \cdot 3$ .

$e = - \frac{1}{3} - (\frac{1}{3 (3)} (3 \cdot - 1))$

Cancel the common factor.

$e = - \frac{1}{3} - (\frac{1}{3 \cdot 3} (3 \cdot - 1))$

Rewrite the expression.

$e = - \frac{1}{3} - (\frac{1}{3} \cdot - 1)$

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

$e = - \frac{1}{3} - \frac{- 1}{3}$

Move the negative in front of the fraction.

$e = - \frac{1}{3} - - \frac{1}{3}$

Multiply $- - \frac{1}{3}$ .

Multiply $- 1$ by $- 1$ .

$e = - \frac{1}{3} + 1 (\frac{1}{3})$

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

$e = - \frac{1}{3} + \frac{1}{3}$

Combine the numerators over the common denominator.

$e = \frac{- 1 + 1}{3}$

Add $- 1$ and $1$ .

$e = \frac{0}{3}$

Divide $0$ by $3$ .

$e = 0$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $- \frac{1}{12} {(y - 2)}^{2} + 0$ .

$- \frac{1}{12} {(y - 2)}^{2} + 0$

Set $x$ equal to the new right side.

$x = - \frac{1}{12} \cdot {(y - 2)}^{2} + 0$

Step 2

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

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

$h = 0$

$k = 2$

Step 3

Find the vertex $(h, k)$ .

$(0, 2)$

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.

Cancel the common factor of $1$ and $- 1$ .

Rewrite $1$ as $- 1 (- 1)$ .

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

Move the negative in front of the fraction.

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

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 $- (1 \cdot 3)$ .

Multiply $3$ by $1$ .

$- 1 \cdot 3$

Multiply $- 1$ by $3$ .

$- 3$

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.

$x = 3$

Step 6

Find the Directrix y^2-4y+12x+4=0

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