Which equation represents a parabola with a focus at (0,-2) and a directrix of y=6?
Accepted Solution
A:
The focus is:
[tex]F(0,-2)[/tex]
Given that the directrix is parallel to the x-axis, then the ordinary equation is given by:
[tex](x-h)^{2} = 4p(y-k)[/tex]
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:
[tex]k = \frac{6-2}{2}[/tex], 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: [tex]p\ \textless \ 0[/tex]