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.