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PLSSS HELP!! I'LL GIVE BRAINLIEST IF RIGHT Which is the equation of a parabola with a directrix at y = 2 and a focus at (5, 0). A) y = one fourth(x − 5)2 + 1 B) y = one fourth(x + 1)2 − 5 C) y = −one fourth(x + 1)2 − 5 d) y = −one fourth(x − 5)2 + 1

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Answer:

The answer is below

Step-by-step explanation:

A parabola is the locus of a point such that its distance from a given line (directrix) and a fixed point (focus) is always the same.

The equation of a parabola with a horizontal line as directrix is given by:

(x - h)² = 4p(y - k)

where y = k - p is the directrix and the focus = (h, k + p)

Given that parabola has a directrix at y = 2 and a focus at (5, 0)

Therefore:

y = k - p

k - p = 2          (1)

focus = (h, k + p) = (5, 0)

h = 5, k + p = 0

k + p = 0         (2)

Solve equation 1 and 2 simultaneously by adding both equations together:

2k = 2

k = 1

Substitute k = 1 in k + p = 0 to find p:

1 + p = 0

p = -1

Therefore h = 5, p = -1 and k = 1

Substitute the values of h, k and p into the equation of a parabola:

(x - h)² = 4p(y - k)

(x - 5)² = 4(-1)(y - 1)

(x - 5)² = -4(y - 1)

x² - 10x + 25 = -4y + 4

-4y = x² - 10x + 21

4y = 10x - x² - 21

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