# Parabola - vertex, focus, directrix

• May 7th 2010, 06:53 AM
aspotonthewall
Parabola - vertex, focus, directrix
The equation below represents a parabola:

$y^2+2y+1=16x-96$

Find:

1) the coordinates of the vertex (xc,yc)
2) the coordinates of the focus (xf,yf)
3) and the equation of the directrix

I don't know where to even begin to solve this.
• May 7th 2010, 07:04 AM
Quote:

Originally Posted by aspotonthewall
The equation below represents a parabola:

$y^2+2y+1=16x-96$

Find:

1) the coordinates of the vertex (xc,yc)
2) the coordinates of the focus (xf,yf)
3) and the equation of the directrix

I don't know where to even begin to solve this.

firstly , put it in the form of its general equation , y^2=4ax

(y+1)^2=4(4)(x-6)

The vertex is (-1,6)

focus=(a,0) where a=4

and the equation of directric is x=-a