Need help with general parabolas (1 Viewer)

[kisschasysunday]

Ok..I have been given a question asking me to find the coordinates of the vertex & focus, as well as the equation of the directrix, of the parabola

x2- 6x- 12y+ 33=0 (the first x being squared...sorry no idea when it comes to html)​

I can do these questions when asked the other way around..as in I can work from the focus and vertex to find the equation of the parabola. I'm completely lost with this question though. I don't know what form it would be, I can't sketch it obviously, and I would reeeaaaallllly appreciate if anyone could give me any help whatsoever. Feel free to add me on msn if that would be easier for you: iamqueen90 @hotmail.com
Thanks!
-Megan

 

[rama_v]

Ok..I have been given a question asking me to find the coordinates of the vertex & focus, as well as the equation of the directrix, of the parabola

x2- 6x- 12y+ 33=0 (the first x being squared...sorry no idea when it comes to html)​

I can do these questions when asked the other way around..as in I can work from the focus and vertex to find the equation of the parabola. I'm completely lost with this question though. I don't know what form it would be, I can't sketch it obviously, and I would reeeaaaallllly appreciate if anyone could give me any help whatsoever. Feel free to add me on msn if that would be easier for you: iamqueen90 @hotmail.com
Thanks!
-Megan

Just use completing the square for these questions, i.e.

x² - 6x + (6/2)² - (6/2)² -12y+33=0
(x-3)²-9+33-12y=0
(x-3)² = 12y - 24
(x-3)²= 12(y - 2) so 4a=12 therefore a =3
and vertex is (3, 2)
go from there

 

[acmilan]

You need to complete the square to get it into the more traditional form:

x²- 6x- 12y+ 33=0
x²- 6x + 9 - 12y+ 24 = 0
(x - 3)² - 12y + 24 = 0
(x - 3)² = 12y - 24
(x - 3)² = 12(y - 2)

There you can now read straight off that. Vertex is (3,2), focal length is 3 (12/4) and hence focus is (3,5)

edit: rama_v beat me to it

 

[kisschasysunday]

You are both geniuses (geniusii?) and thankyou so much!