Find the Directrix 1/3x^2+y=0

$\frac{1}{3} x^{2} + y = 0$

Step 1

Rewrite the equation in vertex form.

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

Combine $\frac{1}{3}$ and $x^{2}$ .

$\frac{x^{2}}{3} + y = 0$

Subtract $\frac{x^{2}}{3}$ from both sides of the equation.

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

Complete the square for $- \frac{x^{2}}{3}$ .

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

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

$b = 0$

$c = 0$

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

$d = \frac{0}{2 (- \frac{1}{3})}$

Simplify the right side.

Cancel the common factor of $0$ and $2$ .

Factor $2$ out of $0$ .

$d = \frac{2 (0)}{2 (- \frac{1}{3})}$

Cancel the common factors.

Cancel the common factor.

$d = \frac{2 \cdot 0}{2 (- \frac{1}{3})}$

Rewrite the expression.

$d = \frac{0}{- \frac{1}{3}}$

Multiply the numerator by the reciprocal of the denominator.

$d = 0 (- 1 \cdot 3)$

Multiply $0 (- 1 \cdot 3)$ .

Multiply $- 1$ by $3$ .

$d = 0 \cdot - 3$

Multiply $0$ by $- 3$ .

$d = 0$

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

$e = 0 - \frac{0^{2}}{4 (- \frac{1}{3})}$

Simplify the right side.

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$e = 0 - \frac{0}{4 (- \frac{1}{3})}$

Simplify the denominator.

Multiply $- 1$ by $4$ .

$e = 0 - \frac{0}{- 4 (\frac{1}{3})}$

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

$e = 0 - \frac{0}{\frac{- 4}{3}}$

Move the negative in front of the fraction.

$e = 0 - \frac{0}{- \frac{4}{3}}$

Multiply the numerator by the reciprocal of the denominator.

$e = 0 - (0 (- \frac{3}{4}))$

Multiply $0 (- \frac{3}{4})$ .

Multiply $- 1$ by $0$ .

$e = 0 - (0 (\frac{3}{4}))$

Multiply $0$ by $\frac{3}{4}$ .

$e = 0 - 0$

Multiply $- 1$ by $0$ .

$e = 0 + 0$

Add $0$ and $0$ .

$e = 0$

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

$- \frac{1}{3} x^{2}$

Set $y$ equal to the new right side.

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

Step 2

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

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

$h = 0$

$k = 0$

Step 3

Find the vertex $(h, k)$ .

$(0, 0)$

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.

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

Simplify.

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

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

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

Move the negative in front of the fraction.

$- \frac{1}{4 (\frac{1}{3})}$

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

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

Multiply the numerator by the reciprocal of the denominator.

$- (1 (\frac{3}{4}))$

Multiply $\frac{3}{4}$ by $1$ .

$- \frac{3}{4}$

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.

$y = \frac{3}{4}$

Step 6

Find the Directrix 1/3x^2+y=0

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