What is the equation in standard form of the parabola with a focus at (51,14) and a directrix of y= 16?

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Answer 1 · 2017-04-22T18:24:10

#y = 1/4x^2 -51/2x + 2661/4#

Because the directrix is a horizontal line, we know that is x coordinate of the vertex is the same as the x coordinate of the focus, #h = 51#.

The y coordinate of the vertex is halfway from the directrix to the focus, #k = 15#

The signed distance from the vertex to the focus is, #f = 1#

Using the vertex form of a parabola of this type:

#y = a(x - h)^2 + k#

Substitute the values for h and k:

#y = a(x - 51)^2 + 15#

Use the formula #a = 1/(4f)#

#a = 1/(4(1))#

#a = 1/4#

Substitute this into the equation:

#y = 1/4(x - 51)^2 + 15#

Expand the square:

#y = 1/4(x^2 -102x + 2601) + 15#

Distribute the #1/4#:

#y = 1/4x^2 -51/2x + 2601/4 + 15#

Combine the constant terms:

#y = 1/4x^2 -51/2x + 2661/4" "larr# This is the standard form.