Asked by doug
determine vertex, focus and directrix of parabola. then graph the parabola.
first equation:
y^2 -2x-8=0
second equation:
y^2 -4y -12x= 8
first equation:
y^2 -2x-8=0
second equation:
y^2 -4y -12x= 8
Answers
Answered by
drwls
We can't draw graphs for you here. I sugggest you do that yourself
Rewrite them in the form
x = a(y-b)^2 + c
When you have done that, the vertex will be at x=c, y=b. The focus will be located at y=b, x = c + p, and the directrix will be the line x = c - p
The value of p is
p = 1/(4a)
For your first equation,
x = y^2/2 -4
so a = 1/2, b = 0 and c = -4
Rewrite them in the form
x = a(y-b)^2 + c
When you have done that, the vertex will be at x=c, y=b. The focus will be located at y=b, x = c + p, and the directrix will be the line x = c - p
The value of p is
p = 1/(4a)
For your first equation,
x = y^2/2 -4
so a = 1/2, b = 0 and c = -4
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