# What is the equation of the parabola with a vertex at (0,0) and a focus at (0,-3)?

Jul 21, 2017

$y = - \frac{1}{12} {x}^{2}$

#### Explanation:

As vertex is $\left(0 , 0\right)$ and focus is $\left(0 , - 3\right)$, the equation is of the form

$y = - a {x}^{2}$ and directrix is $y = 3$

As every point on parabola is equidistant from focus and directrix, and the equation would be

${x}^{2} + {\left(y + 3\right)}^{2} = {\left(y - 3\right)}^{2}$

or ${x}^{2} + {y}^{2} + 6 y + 9 = {y}^{2} - 6 y + 9$

or $12 y = - {x}^{2}$

or $y = - \frac{1}{12} {x}^{2}$

graph{(12y+x^2)(x^2+y^2-0.02)(x^2+(y+3)^2-0.01)(y-3)=0 [-10, 10, -5, 5]}

Jul 21, 2017

$y = - {x}^{2} / 12$

#### Explanation:

Given -
Vertex $= \left(0 , 0\right)$
Focus $= \left(0 , - 3\right)$

This parabola opens down. Look at the graph - So the equation of the parabola in such cases is

${x}^{2} = - 4 a y$

Where $a = 3$
It is the distance from the vertex to focus.

Then -

${x}^{2} = - 4 \times 3 \times y$
${x}^{2} = - 12 y$
$- 12 y = {x}^{2}$
$y = - {x}^{2} / 12$