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?

Question

7452 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-07T12:13:04

The equation of parabola is $y=-1/24x^2$ .

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

between focus and directrix. Therefore vertex is at $(0,(-6+6)/2)$

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

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

So the equation of parabola is $y=a(x-0)^2+0 or y =ax^2$ .

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

$d = 1/(4|a|) :. 6 = 1/(4|a|) or |a|= 1/(4*6)=1/24$ . Here the directrix

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

negative $:a=-1/24$ . Hence the equation of parabola is

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