Find the Focus x-1=1/12(y-3)^2

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

Step 1

Rewrite the equation in vertex form.

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

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

Rewrite.

$x - 1 = 0 + 0 + \frac{1}{12} {(y - 3)}^{2}$

Simplify by adding zeros.

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

Combine $\frac{1}{12}$ and ${(y - 3)}^{2}$ .

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

Add $1$ to both sides of the equation.

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

Reorder terms.

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

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

Simplify the expression.

Simplify each term.

Rewrite ${(y - 3)}^{2}$ as $(y - 3) (y - 3)$ .

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

Expand $(y - 3) (y - 3)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

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

Move $- 3$ to the left of $y$ .

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

Multiply $- 3$ by $- 3$ .

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

Subtract $3 y$ from $- 3 y$ .

$\frac{1}{12} \cdot (y^{2} - 6 y + 9) + 1$

Apply the distributive property.

$\frac{1}{12} y^{2} + \frac{1}{12} (- 6 y) + \frac{1}{12} \cdot 9 + 1$

Simplify.

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

$\frac{y^{2}}{12} + \frac{1}{12} (- 6 y) + \frac{1}{12} \cdot 9 + 1$

Cancel the common factor of $6$ .

Factor $6$ out of $12$ .

$\frac{y^{2}}{12} + \frac{1}{6 (2)} (- 6 y) + \frac{1}{12} \cdot 9 + 1$

Factor $6$ out of $- 6 y$ .

$\frac{y^{2}}{12} + \frac{1}{6 (2)} (6 (- y)) + \frac{1}{12} \cdot 9 + 1$

Cancel the common factor.

$\frac{y^{2}}{12} + \frac{1}{6 \cdot 2} (6 (- y)) + \frac{1}{12} \cdot 9 + 1$

Rewrite the expression.

$\frac{y^{2}}{12} + \frac{1}{2} (- y) + \frac{1}{12} \cdot 9 + 1$

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

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

Cancel the common factor of $3$ .

Factor $3$ out of $12$ .

$\frac{y^{2}}{12} - \frac{y}{2} + \frac{1}{3 (4)} \cdot 9 + 1$

Factor $3$ out of $9$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Combine $\frac{1}{4}$ and $3$ .

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

Write $1$ as a fraction with a common denominator.

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

Combine the numerators over the common denominator.

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

Add $3$ and $4$ .

$\frac{y^{2}}{12} - \frac{y}{2} + \frac{7}{4}$

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

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

$b = - \frac{1}{2}$

$c = \frac{7}{4}$

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

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

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

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

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

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

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

Factor $2$ out of $2$ .

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

Cancel the common factors.

Factor $2$ out of $12$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $6$ by $1$ .

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Factor $2$ out of $6$ .

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

Cancel the common factor.

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

Rewrite the expression.

$d = - 1 \cdot 3$

Multiply $- 1$ by $3$ .

$d = - 3$

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

$e = \frac{7}{4} - \frac{{(- \frac{1}{2})}^{2}}{4 (\frac{1}{12})}$

Simplify the right side.

Simplify each term.

Simplify the numerator.

Apply the product rule to $- \frac{1}{2}$ .

$e = \frac{7}{4} - \frac{{(- 1)}^{2} {(\frac{1}{2})}^{2}}{4 (\frac{1}{12})}$

Raise $- 1$ to the power of $2$ .

$e = \frac{7}{4} - \frac{1 {(\frac{1}{2})}^{2}}{4 (\frac{1}{12})}$

Apply the product rule to $\frac{1}{2}$ .

$e = \frac{7}{4} - \frac{1 \frac{1^{2}}{2^{2}}}{4 (\frac{1}{12})}$

One to any power is one.

$e = \frac{7}{4} - \frac{1 \frac{1}{2^{2}}}{4 (\frac{1}{12})}$

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

$e = \frac{7}{4} - \frac{1 (\frac{1}{4})}{4 (\frac{1}{12})}$

Multiply $\frac{1}{4}$ by $1$ .

$e = \frac{7}{4} - \frac{\frac{1}{4}}{4 (\frac{1}{12})}$

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

$e = \frac{7}{4} - \frac{\frac{1}{4}}{\frac{4}{12}}$

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

Factor $4$ out of $4$ .

$e = \frac{7}{4} - \frac{\frac{1}{4}}{\frac{4 (1)}{12}}$

Cancel the common factors.

Factor $4$ out of $12$ .

$e = \frac{7}{4} - \frac{\frac{1}{4}}{\frac{4 \cdot 1}{4 \cdot 3}}$

Cancel the common factor.

$e = \frac{7}{4} - \frac{\frac{1}{4}}{\frac{4 \cdot 1}{4 \cdot 3}}$

Rewrite the expression.

$e = \frac{7}{4} - \frac{\frac{1}{4}}{\frac{1}{3}}$

Multiply the numerator by the reciprocal of the denominator.

$e = \frac{7}{4} - (\frac{1}{4} \cdot 3)$

Combine $\frac{1}{4}$ and $3$ .

$e = \frac{7}{4} - \frac{3}{4}$

Combine the numerators over the common denominator.

$e = \frac{7 - 3}{4}$

Subtract $3$ from $7$ .

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

Divide $4$ by $4$ .

$e = 1$

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

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

Set $x$ equal to the new right side.

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

Step 2

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

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

$h = 1$

$k = 3$

Step 3

Find the vertex $(h, k)$ .

$(1, 3)$

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.

$(4, 3)$

Step 6

Find the Focus x-1=1/12(y-3)^2

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