How do you find the standard form of the equation of the parabola with a focus at (0, -6) and a directrix at y = 6?

Mar 7, 2018

The equation of parabola is $y = - \frac{1}{24} {x}^{2}$.

Explanation:

Focus is at $\left(0 , - 6\right)$and directrix is $y = 6$. Vertex is at midway

between focus and directrix. Therefore vertex is at $\left(0 , \frac{- 6 + 6}{2}\right)$

or at $\left(0 , 0\right)$ . The vertex form of equation of parabola is

y=a(x-h)^2+k ; (h.k) ; being vertex. $h = 0 \mathmr{and} k = 0$

So the equation of parabola is $y = a {\left(x - 0\right)}^{2} + 0 \mathmr{and} y = a {x}^{2}$.

Distance of vertex from directrix is $d = 6 - 0 = 6$, we know

$d = \frac{1}{4 | a |} \therefore 6 = \frac{1}{4 | a |} \mathmr{and} | a | = \frac{1}{4 \cdot 6} = \frac{1}{24}$. Here the directrix

is above the vertex , so parabola opens downward and $a$ is

negative $: a = - \frac{1}{24}$. Hence the equation of parabola is

$y = - \frac{1}{24} {x}^{2}$.
graph{-1/24 x^2 [-160, 160, -80, 80]} [Ans]