Posted by Ama on Tuesday, November 13, 2012 at 7:27am.
This is an X-parabola.
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 = 0-X1 = 1/4a.
0-X1 = 1/(1/2) = 2.
X1 = -2.
VF = X2-0 = 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 = 0-Y1 = 8/2 = 4.
Y1 = -4.
FB = Y2-0 = 8/2 = 4
Y2 = 4.
precalculus - Find the focus, directrix, and focal diameter of the parabola. x2...
precalculus - Find the focus, directrix, and focal diameter of the parabola. 9x...
precalculus - Find the focus, directrix, and focal diameter of the parabola. x...
algebra - 6. Find the equation of each parabola described below. a) parabola ...
maths - Given that the equation of the parabola is 5y^2 + 24x = 0. Find (1)The ...
pre cal - What is the center of the conic whose equation is x^2 + 2y^2 - 6x + ...
Pre Cal - write the equations of the parabola, the directrix, and the axis of ...
Algebra: Parabolas - I have a math problem that requires me to find the focal ...
Math - Find the vertex, focus, and directrix of the parabola. x^2 - 2x + 8y + 9...
Pre Cal Check - focus: (0,p) vertex: if not given, then it is assumed (0,0) ...
For Further Reading