Question

Find the standard form of the equation of the parabola with a focus at (0, -2) and a directrix at y = 2.

Answer 1

Hello,
The parabola having like focus (0,p/2) and as directrix y=-p/2 has as equation x²=2py

Here -p/2=2==>p=-4

x²=-8y is the equation.

Answer 2

`y=-(1)/(8)x^2`

Explanation:

Find the standard form of the equation of the parabola with a focus at (0, -2) and a directrix at y = 2

The distance between the focust and the directrix is the value of 2p

Distance beween focus (0,-2) and y=2 is 4

`2p=4, p=2`

The distance between vertex and focus is p that is 2

Focus is at (0,-2) , so the vertex is at (0,0)

General form of equation is

`y-k=-(1)/(4p)(x-h)^2`

where (h,k) is the vertex

Vertex is (0,0) and p = 2

The equation becomes

`y-0=-(1)/(4(2))(x-0)^2`

`y=-(1)/(8)x^2`