Question

How do you find the vertex, focus and directrix of `x^2 = 12y`?

Answer 1

`"vertex=(0;0)"`
`f=-d=1/(4*0.083333)=(0;3)`
Directrix is `y+3=0`

Explanation:

`y=1/12*x^2=0.083333*x^2`
`"vertex=(0;0)"`
`"equation=y=a*x^2"`
`"focus, f=a*(2*f)^2"`
`f=-d=1/(4*a)`
`f=-d=1/(4*0.083333)=3`
Directrix is `y+3=0`
graph{(y-1/12x^2)(y-0x-3)(y-0*x+3)=0 [-10, 10, -4, 6]}