What is the standard form of the parabola with a vertex at (4,0) and a focus at (4,-4)?

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Answer 1 · 2017-06-30T15:52:16

$y = - \frac{1}{16} {(x - 4)}^{2}$ $y = -1/16(x - 4)^2$

The standard form of a parabola is

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

where $(h, k)$ $(h, k)$ is the vertex and $p$ $p$ is the distance from the vertex to the focus (or the distance from the vertex to the directrix).

Since we are given the vertex $(4, 0)$ $(4, 0)$ , we can plug this into our parabola formula.

$y = \frac{1}{4 p} {(x - 4)}^{2} + 0$ $y = 1/(4p)(x - 4)^2 + 0$

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

To help visualize $p$ $p$ , let's plot our given points on a graph.

Desmos

$p$ $p$ , or the distance from the vertex to the focus, is -4. Plug this value into the equation:

$y = \frac{1}{4 (- 4)} {(x - 4)}^{2}$ $y = 1/(4(-4))(x - 4)^2$
$y = - \frac{1}{16} {(x - 4)}^{2}$ $y = -1/16(x - 4)^2$

That's your parabola in standard form!

Desmos