What is the equation in standard form of the parabola with a focus at (2,-4) and a directrix of y= 6?

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Answer 1 · 2016-11-07T15:54:40

$y = - \frac{1}{20} x^{2} + \frac{1}{5} x + \frac{4}{5}$ $y = -1/20x^2 + 1/5x + 4/5$

The general form for the equation of a horizontal directrix is $y = k - f$ $y = k - f$

The focus of a parabola with a horizontal directrix is of the general form $(h, k + f)$ $(h, k + f)$

Therefore, we can write these 3 equations that will help us:

h = 2

-4 = k + f

6 = k - f

solve the last two equation for k and f:

$k = 1$ $k = 1$
$f = - 5$ $f = -5$

The vertex form of the equation of this type of parabola is:

$y = a {(x - h)}^{2} + k$ $y = a(x - h)^2 + k$

Because we are not given the value of a, substitute $\frac{1}{4 f}$ $1/(4f)$ for $a$ $a$ :

$y = \frac{1}{4 f} {(x - h)}^{2} + k$ $y = 1/(4f)(x - h)^2 + k$

Substitute our known values into the above equation:

$y = \frac{1}{4 (- 5)} {(x - 2)}^{2} + 1$ $y = 1/(4(-5))(x - 2)^2 + 1$

Expand the square:

$y = - \frac{1}{20} (x^{2} - 4 x + 4) + 1$ $y = -1/20(x^2 - 4x+ 4) + 1$

Distribute $- \frac{1}{20}$ $-1/20$ :

$y = - \frac{1}{20} x^{2} + \frac{1}{5} x - \frac{1}{5} + 1$ $y = -1/20x^2 + 1/5x - 1/5 + 1$

Combine the constants:

$y = - \frac{1}{20} x^{2} + \frac{1}{5} x + \frac{4}{5}$ $y = -1/20x^2 + 1/5x + 4/5$