y^2 + 2 y = 4 x - 33
y^2 + 2 y + 1 = 4 x - 32
(y+1)^2 = 4 (x-8)
(8,-1) is the vertex etc.
opens right, a = 4
vertex to focus = 4 so focus at (12,-1)
vertex to directrix = 4 so directrix at x = 4
Find the vertex, focus and directrix of the parabola.
4x - y^2 - 2y - 33 = 0
y^2+2y+1 = -4x+33+1
(y+1)^2 = -4(x+8.5)
vertex:(-1,-8.5)
focus:(-2,-8.5)
Directrix: x=0
Is this correct?
1 answer