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

Verified Answer :

caylus

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:

\(y=-\frac{1}{8}x^2\)

Step-by-step 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

\(y-k=-\frac{1}{4p}(x-h)^2\)

where (h,k) is the vertex

Vertex is (0,0) and p = 2

The equation becomes

\(y-0=-\frac{1}{4(2)}(x-0)^2\)

\(y=-\frac{1}{8}x^2\)