precalculus
posted by Ama .
Find the focus, directrix, and focal diameter of the parabola.
x2 = 5y
focus (x, y) = 0, 5/4
directrix:y=5/4 focal diameter=5
Sketch its graph?
my answers seem to be correct but i cannnot sketch the graph correctly.

X^2 = 5y. Yparabola.
Y = (1/5)x^2
a = 1/5, 4a = 4/5, 1/4a = 5/4.
h = Xv = b/2a = 0/(2/5) = 0.
k = Yv = (1/5)*0^2 = 0.
V(h,k) = V(0,0).
D(h,Y1), V(h,k), F(h,Y2). Ver. Line.
D(0,Y1), V(0,0), F(0,Y2).
DV = 0Y1 = 1/4a
0Y1 = 1/4a = 5/4.
Y1 = 5/4.
VF = Y20 = 1/4a.
Y20 = 1/4a = 5/4
Y2 = 5/4.
Focal Dia. = 1/a = 1/(1/5) = 5.
A(X1,5/4), F(0,5/4), B(X2,5/4). Hor line
AF = 0X1 = 5/2
0X1 = 5/2
X1 = 5/2.
FB = X20 = 5/2
X2 = 5/2.
Your answers are correct!
To graph the parabola, select values of
X below and above h(Xv); and calculate
the corresponding value of Y in each
case.
(x,Y). Y = (1/5)x^2.
(3,9/5).
(2,4/5).
(1,1/5).
V(0,0).
(1,1/5).
(2,4/5).
(3,9/5).