Parabolas, conic sections

[lillie5455]

In Algebra II, we will have an equation like y^2=8x and to solve it, I need to find the focus and dirictrix when the vertex is (0,0). To do this, my book says to us the equation y^2=4px, so 4p=8, and p=2. This makes the focus (2,0) and the directrix is x=-2. I understand how to write the equation when the focus is (0,0),but I don't understand how to find all of this when the vertex is at a point like (3,-6). My book wants me to write an equation of the parabola with a vertex at (3,-6) and a focus at (3,-4). It gives me an equation to figure it out, which is (x-h)^2=4p(y-k) and (y-k)^2=4p(x-h). I don't know how to decide which one is squared by looking at the given points or how to set it up. Please help. I also do not know how to graph the new equation.[/i]

 

[discreditedvalidity]

Did you try plugging in the x and y values into the equation?

if there is a point (3, -6), plug 3 in for x and -6 in for y, then solve for the missing variable.

 

[galactus]

By the looks of your given coordinates, we can tell your parabola opens upward because the y-coordinate of the focus is above the y-coordinate of the vertex.

Use \(\displaystyle (x-h)^{2}=4p(y-k)\)

p is the distance from the focus to the vertex. and (h,k) are the coordinates of the vertex

 

[lillie5455]

So my equation is (x-3)^2=4p(y+6)?
If so what is p?

 

[lillie5455]

So my equation is (x-3)^2=8(y+6)?

 

[galactus]

Yes, very good. Now, can you solve for y?. It isn't very difficult.

 

[lillie5455]

(x-3)^2=8y+48.
Thank you very much. I was really struggling.