Algebra Examples

$(0, 0)$

$y = - 6$

Step 1

Since the directrix is vertical, use the equation of a parabola that opens up or down.

${(x - h)}^{2} = 4 p (y - k)$

Step 2

Find the vertex.

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Step 2.1

The vertex

$(h, k)$ is halfway between the directrix and focus. Find the

$y$ coordinate of the vertex using the formula

$y = \frac{y coordinate of focus + directrix}{2}$ . The

$x$ coordinate will be the same as the

$x$ coordinate of the focus.

$(0, \frac{0 - 6}{2})$

Step 2.2

Simplify the vertex.

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Step 2.2.1

Cancel the common factor of

$0 - 6$ and

$2$ .

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Step 2.2.1.1

Factor

$2$ out of

$0$ .

$(0, \frac{2 \cdot 0 - 6}{2})$

Step 2.2.1.2

Factor

$2$ out of

$- 6$ .

$(0, \frac{2 \cdot 0 + 2 \cdot - 3}{2})$

Step 2.2.1.3

Factor

$2$ out of

$2 \cdot 0 + 2 \cdot - 3$ .

$(0, \frac{2 \cdot (0 - 3)}{2})$

Step 2.2.1.4

Cancel the common factors.

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Step 2.2.1.4.1

Factor

$2$ out of

$2$ .

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

Step 2.2.1.4.2

Cancel the common factor.

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

Step 2.2.1.4.3

Rewrite the expression.

$(0, \frac{0 - 3}{1})$

Step 2.2.1.4.4

Divide

$0 - 3$ by

$1$ .

$(0, 0 - 3)$

Step 2.2.2

Subtract

$3$ from

$0$ .

$(0, - 3)$

Step 3

Find the distance from the focus to the vertex.

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Step 3.1

The distance from the focus to the vertex and from the vertex to the directrix is

$| p |$ . Subtract the

$y$ coordinate of the vertex from the

$y$ coordinate of the focus to find

$p$ .

$p = 0 + 3$

Step 3.2

Add

$0$ and

$3$ .

$p = 3$

Step 4

Substitute in the known values for the variables into the equation

${(x - h)}^{2} = 4 p (y - k)$ .

${(x - 0)}^{2} = 4 (3) (y + 3)$

Step 5

Simplify.

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

Step 6