Thread: pre-calc--parabolas

1. pre-calc--parabolas

Hey, I have a huge test tomorrow and I haven't been there in awhile so I don't know anything! Here's a typical problem that has been troubling me...

Find the vertex, focus, and directrix of the parabola

x^2 + 4x + 6y -2 = 0

y^2 - 4x -4 = 0

Can anyone help me? Thanks!

2. Originally Posted by simplytrite
Hey, I have a huge test tomorrow and I haven't been there in awhile so I don't know anything! Here's a typical problem that has been troubling me...

Find the vertex, focus, and directrix of the parabola

x^2 + 4x + 6y -2 = 0

y^2 - 4x -4 = 0

Can anyone help me? Thanks!
Solve each for the stadard form:
(x - h)^2 = 4p(y - k) <-- vertical parabola
(y - k)^2 = 4p(x - h) <-- horizontal parabola

x^2 + 4x + 6y - 2 = 0
(x^2 + 4x + 4) = -6y + 2 + 4 = -6y + 6
(x + 2)^2 = -6(y - 1) <-- vertical parabola

The focal length is: 4p = -6 --> p = -3/2 <-- this is the distance from the vertex to the directrix line and to the focus. Since this parabola is VERTICAL, the focus and directrix are ABOVE and BELOW the vertex.

The vertex is: (h,k) = (-2,1)
The focus is "p" units above the vertex: (-2,1 + p) = (-2,1 - 3/2) = (-2,-1/2)
The directrix is "p" units below the vertex: y = 1 - p = 1 - (-3/2) = 5/2

3. Originally Posted by simplytrite
Hey, I have a huge test tomorrow and I haven't been there in awhile so I don't know anything! Here's a typical problem that has been troubling me...

Find the vertex, focus, and directrix of the parabola

x^2 + 4x + 6y -2 = 0

y^2 - 4x -4 = 0

Can anyone help me? Thanks!
y^2 - 4x - 4 = 0
y^2 = 4(x + 1) <-- horizontal parabola

The focal length is: 4p = 4 --> p = 1 <-- Since this parabola is HORIZONTAL, the focus and directrix are to the RIGHT and LEFT of the vertex.

The vertex is: (h,k) = (-1,0)
The focus is "p" units to the right of the vertex: (-1 + p,0) = (-1 + 1,0) = (0,0)
The directrix is "p" units to the left of the vertex: x = -1 - p = -1 - 1 = -2