Precalculus Examples

$(0, - 8)$ $y = 8$

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.

$(0, \frac{- 8 + 8}{2})$

Step 2.2

Simplify the vertex.

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Step 2.2.1

Cancel the common factor of $- 8 + 8$ and $2$ .

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Step 2.2.1.1

Factor $2$ out of $- 8$ .

$(0, \frac{2 \cdot - 4 + 8}{2})$

Step 2.2.1.2

Factor $2$ out of $8$ .

$(0, \frac{2 \cdot - 4 + 2 \cdot 4}{2})$

Step 2.2.1.3

Factor $2$ out of $2 \cdot - 4 + 2 \cdot 4$ .

$(0, \frac{2 \cdot (- 4 + 4)}{2})$

Step 2.2.1.4

Cancel the common factors.

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Step 2.2.1.4.1

Factor $2$ out of $2$ .

$(0, \frac{2 \cdot (- 4 + 4)}{2 (1)})$

Step 2.2.1.4.2

Cancel the common factor.

$(0, \frac{2 \cdot (- 4 + 4)}{2 \cdot 1})$

Step 2.2.1.4.3

Rewrite the expression.

$(0, \frac{- 4 + 4}{1})$

Step 2.2.1.4.4

Divide $- 4 + 4$ by $1$ .

$(0, - 4 + 4)$

Step 2.2.2

Add $- 4$ and $4$ .

$(0, 0)$

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 $p$ .

$p = - 8 - 0$

Step 3.2

Subtract $0$ from $- 8$ .

$p = - 8$

Step 4

Substitute in the known values for the variables into the equation ${(x - h)}^{2} = 4 p (y - k)$ .

${(x - 0)}^{2} = 4 (- 8) (y - 0)$

Step 5

Simplify.

$x^{2} = - 32 y$

Step 6