precalculus
posted by Ama .
Find the focus, directrix, and focal diameter of the parabola.
y2 = 8x

This is an Xparabola.
Y^2 = 8x.
X = (1/8)y^2
k = Yv = b/2a = 0/(1/4) = 0
h = Xv = (1/8*0^2 = 0.
V(h,k) = V(0,0).
D(X1,K), V(h,k), F(X2,K).
D(X1,0), V(0,0), F(X2,0).
DV = 0X1 = 1/4a.
0X1 = 1/(1/2) = 2.
X1 = 2.
VF = X20 = 1/4a
X2 = 1/(1/2) = 2
Focal Dia.(Ver. line).
A(2,Y1), F(2,0), B(2,Y2).
a = 1/8
1/a = 8 = Focal Dia.
AF = 0Y1 = 8/2 = 4.
Y1 = 4.
FB = Y20 = 8/2 = 4
Y2 = 4.