Find the Focus (y+2)^2=12(x+3)

${(y + 2)}^{2} = 12 (x + 3)$

Step 1

Rewrite the equation in vertex form.

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

Rewrite the equation as $12 (x + 3) = {(y + 2)}^{2}$ .

$12 (x + 3) = {(y + 2)}^{2}$

Divide each term in $12 (x + 3) = {(y + 2)}^{2}$ by $12$ and simplify.

Divide each term in $12 (x + 3) = {(y + 2)}^{2}$ by $12$ .

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

Simplify the left side.

Cancel the common factor of $12$ .

Cancel the common factor.

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

Divide $x + 3$ by $1$ .

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

Subtract $3$ from both sides of the equation.

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

Reorder terms.

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

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

Simplify the expression.

Simplify each term.

Rewrite ${(y + 2)}^{2}$ as $(y + 2) (y + 2)$ .

$\frac{1}{12} \cdot ((y + 2) (y + 2)) - 3$

Expand $(y + 2) (y + 2)$ using the FOIL Method.

Apply the distributive property.

$\frac{1}{12} \cdot (y (y + 2) + 2 (y + 2)) - 3$

Apply the distributive property.

$\frac{1}{12} \cdot (y \cdot y + y \cdot 2 + 2 (y + 2)) - 3$

Apply the distributive property.

$\frac{1}{12} \cdot (y \cdot y + y \cdot 2 + 2 y + 2 \cdot 2) - 3$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

$\frac{1}{12} \cdot (y^{2} + y \cdot 2 + 2 y + 2 \cdot 2) - 3$

Move $2$ to the left of $y$ .

$\frac{1}{12} \cdot (y^{2} + 2 \cdot y + 2 y + 2 \cdot 2) - 3$

Multiply $2$ by $2$ .

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

Add $2 y$ and $2 y$ .

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

Apply the distributive property.

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

Simplify.

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

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

Cancel the common factor of $4$ .

Factor $4$ out of $12$ .

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

Factor $4$ out of $4 y$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Cancel the common factor of $4$ .

Factor $4$ out of $12$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

To write $- 3$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

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

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 3$ by $3$ .

$\frac{y^{2}}{12} + \frac{y}{3} + \frac{1 - 9}{3}$

Subtract $9$ from $1$ .

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

Move the negative in front of the fraction.

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

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

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

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

$c = - \frac{8}{3}$

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.

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

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

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

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

Factor $2$ out of $2$ .

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

Cancel the common factors.

Factor $2$ out of $12$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $6$ by $1$ .

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

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

$e = - \frac{8}{3} - \frac{{(\frac{1}{3})}^{2}}{4 (\frac{1}{12})}$

Simplify the right side.

Simplify each term.

Simplify the numerator.

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

$e = - \frac{8}{3} - \frac{\frac{1^{2}}{3^{2}}}{4 (\frac{1}{12})}$

One to any power is one.

$e = - \frac{8}{3} - \frac{\frac{1}{3^{2}}}{4 (\frac{1}{12})}$

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

$e = - \frac{8}{3} - \frac{\frac{1}{9}}{4 (\frac{1}{12})}$

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

$e = - \frac{8}{3} - \frac{\frac{1}{9}}{\frac{4}{12}}$

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

Factor $4$ out of $4$ .

$e = - \frac{8}{3} - \frac{\frac{1}{9}}{\frac{4 (1)}{12}}$

Cancel the common factors.

Factor $4$ out of $12$ .

$e = - \frac{8}{3} - \frac{\frac{1}{9}}{\frac{4 \cdot 1}{4 \cdot 3}}$

Cancel the common factor.

$e = - \frac{8}{3} - \frac{\frac{1}{9}}{\frac{4 \cdot 1}{4 \cdot 3}}$

Rewrite the expression.

$e = - \frac{8}{3} - \frac{\frac{1}{9}}{\frac{1}{3}}$

Multiply the numerator by the reciprocal of the denominator.

$e = - \frac{8}{3} - (\frac{1}{9} \cdot 3)$

Cancel the common factor of $3$ .

Factor $3$ out of $9$ .

$e = - \frac{8}{3} - (\frac{1}{3 (3)} \cdot 3)$

Cancel the common factor.

$e = - \frac{8}{3} - (\frac{1}{3 \cdot 3} \cdot 3)$

Rewrite the expression.

$e = - \frac{8}{3} - \frac{1}{3}$

Combine the numerators over the common denominator.

$e = \frac{- 8 - 1}{3}$

Subtract $1$ from $- 8$ .

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

Divide $- 9$ by $3$ .

$e = - 3$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $\frac{1}{12} {(y + 2)}^{2} - 3$ .

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

Set $x$ equal to the new right side.

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

Step 2

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

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

$h = - 3$

$k = - 2$

Step 3

Find the vertex $(h, k)$ .

$(- 3, - 2)$

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.

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 focus.

The focus of a parabola can be found by adding $p$ to the x-coordinate $h$ if the parabola opens left or right.

$(h + p, k)$

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

$(0, - 2)$

Step 6

Find the Focus (y+2)^2=12(x+3)

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