Find the Directrix (x-3)^2=20(y-1)

${(x - 3)}^{2} = 20 (y - 1)$

Step 1

Rewrite the equation in vertex form.

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

Rewrite the equation as $20 (y - 1) = {(x - 3)}^{2}$ .

$20 (y - 1) = {(x - 3)}^{2}$

Divide each term in $20 (y - 1) = {(x - 3)}^{2}$ by $20$ and simplify.

Divide each term in $20 (y - 1) = {(x - 3)}^{2}$ by $20$ .

$\frac{20 (y - 1)}{20} = \frac{{(x - 3)}^{2}}{20}$

Simplify the left side.

Cancel the common factor of $20$ .

Cancel the common factor.

$\frac{20 (y - 1)}{20} = \frac{{(x - 3)}^{2}}{20}$

Divide $y - 1$ by $1$ .

$y - 1 = \frac{{(x - 3)}^{2}}{20}$

Add $1$ to both sides of the equation.

$y = \frac{{(x - 3)}^{2}}{20} + 1$

Reorder terms.

$y = \frac{1}{20} \cdot {(x - 3)}^{2} + 1$

Complete the square for $\frac{1}{20} \cdot {(x - 3)}^{2} + 1$ .

Simplify the expression.

Simplify each term.

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

$\frac{1}{20} \cdot ((x - 3) (x - 3)) + 1$

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

Apply the distributive property.

$\frac{1}{20} \cdot (x (x - 3) - 3 (x - 3)) + 1$

Apply the distributive property.

$\frac{1}{20} \cdot (x \cdot x + x \cdot - 3 - 3 (x - 3)) + 1$

Apply the distributive property.

$\frac{1}{20} \cdot (x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3) + 1$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\frac{1}{20} \cdot (x^{2} + x \cdot - 3 - 3 x - 3 \cdot - 3) + 1$

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

$\frac{1}{20} \cdot (x^{2} - 3 \cdot x - 3 x - 3 \cdot - 3) + 1$

Multiply $- 3$ by $- 3$ .

$\frac{1}{20} \cdot (x^{2} - 3 x - 3 x + 9) + 1$

Subtract $3 x$ from $- 3 x$ .

$\frac{1}{20} \cdot (x^{2} - 6 x + 9) + 1$

Apply the distributive property.

$\frac{1}{20} x^{2} + \frac{1}{20} (- 6 x) + \frac{1}{20} \cdot 9 + 1$

Simplify.

Combine $\frac{1}{20}$ and $x^{2}$ .

$\frac{x^{2}}{20} + \frac{1}{20} (- 6 x) + \frac{1}{20} \cdot 9 + 1$

Cancel the common factor of $2$ .

Factor $2$ out of $20$ .

$\frac{x^{2}}{20} + \frac{1}{2 (10)} (- 6 x) + \frac{1}{20} \cdot 9 + 1$

Factor $2$ out of $- 6 x$ .

$\frac{x^{2}}{20} + \frac{1}{2 (10)} (2 (- 3 x)) + \frac{1}{20} \cdot 9 + 1$

Cancel the common factor.

$\frac{x^{2}}{20} + \frac{1}{2 \cdot 10} (2 (- 3 x)) + \frac{1}{20} \cdot 9 + 1$

Rewrite the expression.

$\frac{x^{2}}{20} + \frac{1}{10} (- 3 x) + \frac{1}{20} \cdot 9 + 1$

Combine $- 3$ and $\frac{1}{10}$ .

$\frac{x^{2}}{20} + \frac{- 3}{10} x + \frac{1}{20} \cdot 9 + 1$

Combine $\frac{- 3}{10}$ and $x$ .

$\frac{x^{2}}{20} + \frac{- 3 x}{10} + \frac{1}{20} \cdot 9 + 1$

Combine $\frac{1}{20}$ and $9$ .

$\frac{x^{2}}{20} + \frac{- 3 x}{10} + \frac{9}{20} + 1$

Move the negative in front of the fraction.

$\frac{x^{2}}{20} - \frac{3 x}{10} + \frac{9}{20} + 1$

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

$\frac{x^{2}}{20} - \frac{3 x}{10} + \frac{9}{20} + \frac{20}{20}$

Combine the numerators over the common denominator.

$\frac{x^{2}}{20} - \frac{3 x}{10} + \frac{9 + 20}{20}$

Add $9$ and $20$ .

$\frac{x^{2}}{20} - \frac{3 x}{10} + \frac{29}{20}$

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

$a = \frac{1}{20}$

$b = - \frac{3}{10}$

$c = \frac{29}{20}$

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

$d = \frac{- \frac{3}{10}}{2 (\frac{1}{20})}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$d = - \frac{3}{10} \cdot \frac{1}{2 (\frac{1}{20})}$

Combine $2$ and $\frac{1}{20}$ .

$d = - \frac{3}{10} \cdot \frac{1}{\frac{2}{20}}$

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

Factor $2$ out of $2$ .

$d = - \frac{3}{10} \cdot \frac{1}{\frac{2 (1)}{20}}$

Cancel the common factors.

Factor $2$ out of $20$ .

$d = - \frac{3}{10} \cdot \frac{1}{\frac{2 \cdot 1}{2 \cdot 10}}$

Cancel the common factor.

$d = - \frac{3}{10} \cdot \frac{1}{\frac{2 \cdot 1}{2 \cdot 10}}$

Rewrite the expression.

$d = - \frac{3}{10} \cdot \frac{1}{\frac{1}{10}}$

Multiply the numerator by the reciprocal of the denominator.

$d = - \frac{3}{10} (1 \cdot 10)$

Cancel the common factor of $10$ .

Move the leading negative in $- \frac{3}{10}$ into the numerator.

$d = \frac{- 3}{10} (1 \cdot 10)$

Factor $10$ out of $1 \cdot 10$ .

$d = \frac{- 3}{10} (10 \cdot 1)$

Cancel the common factor.

$d = \frac{- 3}{10} (10 \cdot 1)$

Rewrite the expression.

$d = - 3$

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

$e = \frac{29}{20} - \frac{{(- \frac{3}{10})}^{2}}{4 (\frac{1}{20})}$

Simplify the right side.

Simplify each term.

Simplify the numerator.

Apply the product rule to $- \frac{3}{10}$ .

$e = \frac{29}{20} - \frac{{(- 1)}^{2} {(\frac{3}{10})}^{2}}{4 (\frac{1}{20})}$

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

$e = \frac{29}{20} - \frac{1 {(\frac{3}{10})}^{2}}{4 (\frac{1}{20})}$

Apply the product rule to $\frac{3}{10}$ .

$e = \frac{29}{20} - \frac{1 \frac{3^{2}}{10^{2}}}{4 (\frac{1}{20})}$

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

$e = \frac{29}{20} - \frac{1 \frac{9}{10^{2}}}{4 (\frac{1}{20})}$

Raise $10$ to the power of $2$ .

$e = \frac{29}{20} - \frac{1 (\frac{9}{100})}{4 (\frac{1}{20})}$

Multiply $\frac{9}{100}$ by $1$ .

$e = \frac{29}{20} - \frac{\frac{9}{100}}{4 (\frac{1}{20})}$

Combine $4$ and $\frac{1}{20}$ .

$e = \frac{29}{20} - \frac{\frac{9}{100}}{\frac{4}{20}}$

Cancel the common factor of $4$ and $20$ .

Factor $4$ out of $4$ .

$e = \frac{29}{20} - \frac{\frac{9}{100}}{\frac{4 (1)}{20}}$

Cancel the common factors.

Factor $4$ out of $20$ .

$e = \frac{29}{20} - \frac{\frac{9}{100}}{\frac{4 \cdot 1}{4 \cdot 5}}$

Cancel the common factor.

$e = \frac{29}{20} - \frac{\frac{9}{100}}{\frac{4 \cdot 1}{4 \cdot 5}}$

Rewrite the expression.

$e = \frac{29}{20} - \frac{\frac{9}{100}}{\frac{1}{5}}$

Multiply the numerator by the reciprocal of the denominator.

$e = \frac{29}{20} - (\frac{9}{100} \cdot 5)$

Cancel the common factor of $5$ .

Factor $5$ out of $100$ .

$e = \frac{29}{20} - (\frac{9}{5 (20)} \cdot 5)$

Cancel the common factor.

$e = \frac{29}{20} - (\frac{9}{5 \cdot 20} \cdot 5)$

Rewrite the expression.

$e = \frac{29}{20} - \frac{9}{20}$

Combine the numerators over the common denominator.

$e = \frac{29 - 9}{20}$

Subtract $9$ from $29$ .

$e = \frac{20}{20}$

Divide $20$ by $20$ .

$e = 1$

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $\frac{1}{20} {(x - 3)}^{2} + 1$ .

$\frac{1}{20} {(x - 3)}^{2} + 1$

Set $y$ equal to the new right side.

$y = \frac{1}{20} \cdot {(x - 3)}^{2} + 1$

Step 2

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

$a = \frac{1}{20}$

$h = 3$

$k = 1$

Step 3

Find the vertex $(h, k)$ .

$(3, 1)$

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.

$\frac{1}{4 \cdot \frac{1}{20}}$

Simplify.

Combine $4$ and $\frac{1}{20}$ .

$\frac{1}{\frac{4}{20}}$

Cancel the common factor of $4$ and $20$ .

Factor $4$ out of $4$ .

$\frac{1}{\frac{4 (1)}{20}}$

Cancel the common factors.

Factor $4$ out of $20$ .

$\frac{1}{\frac{4 \cdot 1}{4 \cdot 5}}$

Cancel the common factor.

$\frac{1}{\frac{4 \cdot 1}{4 \cdot 5}}$

Rewrite the expression.

$\frac{1}{\frac{1}{5}}$

Multiply the numerator by the reciprocal of the denominator.

$1 \cdot 5$

Multiply $5$ by $1$ .

$5$

Step 5

Find the directrix.

The directrix of a parabola is the horizontal line found by subtracting $p$ from the y-coordinate $k$ of the vertex if the parabola opens up or down.

$y = k - p$

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

$y = - 4$

Step 6

Find the Directrix (x-3)^2=20(y-1)

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