How to indentify the vertex, focus and directrix
HOME > > How to indentify the vertex, focus and directrix

How to indentify the vertex, focus and directrix

[From: ] [author: ] [Date: 12-12-20] [Hit: ]
.my teacher never taught us but its on the homework >:(Anyway if you could just help me through solving for one of those equations I would appreciate it SOOO much...Thanks!!......
Can you help me identify the vertex, focus and directrix of the graph of each of these equations??
1) y+1=-1/4(x-3)^2
2) x=2y^2
3) y^2-4x-2y=3
I have no clue how to do this..my teacher never taught us but its on the homework >:( Anyway if you could just help me through solving for one of those equations I would appreciate it SOOO much...just figured I'd post them all in case anybody had some serious time on their hands ;)
Thanks!!

-
Some basic equations:

(x-h)^2 = 4*p*(y-k)...for parabolas that open up & down. The vertex is at (h,k). P is positive for opens up, P is negative for opens down. The focus is at (h, k+p), which is "inside" the parabola. The directrix is at y = k-p, which is above or below the parabola (depending on the value of "p")

(y-k)^2 = 4*p*(x-h)...for parabolas that open left & right. The vertex is at (h,k). P is positive for opens right, P is negative for opens left. Focus is at (h+p, k). The directrix is at x = k-p.

For #1:
Vertex (3, -1)

Fits the first case...
multiply both sides by -4 on both sides.
-4(y+1) = (x-3)^2

-4 = 4p
p = -1

Focus is at (2,-1)
Directrix is at x = 4

#2
This one is a little more straight forward.
Vertex at (0,0)

0.5x = y^2

0.5 = 4p
p = 0.125

Focus is at (0.125 , 0)
Directrix is at x = -0.125

#3

You need to complete the square

y^2 - 2y +1 -4x = 3+1
(y-1)^2 = 4x+4
(y-1)^2 = 4(x+1)

so the vertex is at (-1, 1)

4 = 4p
p = 1

so the focus is at (0,1)
the directrix is at x = -2.
1
keywords: and,vertex,indentify,directrix,How,focus,to,the,How to indentify the vertex, focus and directrix
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .