# How do you find the standard form of the equation of the parabola with the given characteristic and vertex at the origin and Focus: (0, −5)?

Jan 7, 2017

The equation is ${x}^{2} = - 20 y$

#### Explanation:

The vertex is $V = \left(0 , 0\right)$

The focus is $F = \left(0 , - 5\right) = \left(0 , - \frac{p}{2}\right)$

So the directrix is $y = 5$

Therefore,

$p = 10$

Any point on the parabola is equidistant from the focus and the directrix

$\left(5 - y\right) = \sqrt{{x}^{2} + {\left(y + 5\right)}^{2}}$

$\cancel{25} + {\cancel{y}}^{2} - 10 y = {x}^{2} + {\cancel{y}}^{2} + 10 y + \cancel{25}$

The equation of the parabola is

${x}^{2} = - 20 y$

graph{(x^2+20y)(y-5)((x)^2+(y+5)^2-0.01)=0 [-12.66, 12.65, -6.33, 6.33]}