Question

Which equation represents a parabola with a focus at (0,-2) and a directrix of y=6?

Answer 1

The focus is:


`F(0,-2)`

Given that the directrix is parallel to the x-axis, then the ordinary equation is given by:


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

We need to find V(h,k), being V the vertex.

We know that these distances are always the same, namely:

│FV│ = │VD│

Being D the directrix. Given that the focus F is on the y-axis and the directrix is parallel to the x-axis, then the vertex V will also be on this axis, so h = 0.

As │FV│ = │VD│, then:

`k = (6-2)/(2)`, that is the middle point of the segment FD, so:

V(0,2)

Now │FV│= │p│= │2-(-2)│=4

Given that the vertex and focus are below the directrix, then the parabola open down, therefore:
`p\ \textless \ 0`

Lastly, the equation is:

`x^(2) = -4(4)(y-2) = -16y+32`

`y = -( x^(2) )/(16) + 2`