Thursday

October 27, 2016
Posted by **Ama** on Tuesday, November 13, 2012 at 7:33am.

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.

- precalculus -
**Henry**, Friday, November 16, 2012 at 12:10pmX^2 = 5y. Y-parabola.

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 = 0-Y1 = 1/4a

0-Y1 = 1/4a = 5/4.

Y1 = -5/4.

VF = Y2-0 = 1/4a.

Y2-0 = 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 = 0-X1 = 5/2

0-X1 = 5/2

X1 = -5/2.

FB = X2-0 = 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).