the parabola x^2 = 4py has
vertex at (0,0)
focus at (0,p)
directrix at y = -p
So. Let's rearrange things a bit.
12y=(x-1)^2-48
(x-1)^2 = 12(y+4)
So, we can see that we now have
vertex at (1,-4)
focus at y=(-4+3) = (1,-1)
directrix at y=(-4-3) or y = -7
to confirm our work, see
https://www.wolframalpha.com/input/?i=parabola+12y%3D(x-1)%5E2-48
The equation of a parabola is 12y=(x-1)^2-48. Identify the vertex, focus, and directrix of the parabola.
2 answers
12y = (x-1)^2 - 48.
Vertex form: Y = a(x-h)^2 + k.
Y = (1/12)(x-1)^2 - 4.
V(h, k) = V(1, -4).
F(h, k+1/(4a)) = F(1, -1).
D(h, k-1/(4a)) = D(1, -7).
Vertex form: Y = a(x-h)^2 + k.
Y = (1/12)(x-1)^2 - 4.
V(h, k) = V(1, -4).
F(h, k+1/(4a)) = F(1, -1).
D(h, k-1/(4a)) = D(1, -7).
1 answer