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Determine the equation of the parabola with vertex
(-8,5) and directrix
y=-3.

Determine the equation of the parabola with vertex $(-8,5)$ and directrix $y=-3$ .

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Write equations of parabolas in vertex form using properties

Full solution

Q. Determine the equation of the parabola with vertex

(-8,5)

and directrix

y=-3

.

Identify Parabola Type: Identify whether the parabola is vertical or horizontal based on the directrix. $\newline$ Directrix: $y = -3$ is a horizontal line. $\newline$ Type of parabola: Vertical
Vertex Form: Identify the vertex form of a vertical parabola. $\newline$ Vertex form of vertical parabola: $y = a(x-h)^2+k$
Determine Direction: Determine if the parabola opens upward or downward. $\newline$ Vertex $(-8,5)$ is above the directrix $y = -3$ . $\newline$ The parabola opens upward.
Calculate Distance: Calculate the distance between the vertex and the directrix. $\newline$ y-value for the vertex: $5$ $\newline$ y-value for the directrix: $-3$ $\newline$ Distance: $|5 - (-3)| = 5 + 3 = 8$
Find Value of a: Determine the value of a. $\newline$ The distance between the vertex and directrix is $8$ , which is equal to $\frac{1}{4a}$ . $\newline$ $8 = \frac{1}{4a}$ $\newline$ Solve for a: $a = \frac{1}{4\times 8}$ $\newline$ $a = \frac{1}{32}$
Write Parabola Equation: Write the equation of the parabola using the vertex form. $\newline$ Substitute $\frac{1}{32}$ for $a$ , $-8$ for $h$ , and $5$ for $k$ . $\newline$ $y = \frac{1}{32} (x - (-8))^2 + 5$ $\newline$ $y = \frac{1}{32} (x + 8)^2 + 5$

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