# Introduction to CONICS

• Dec 6th 2009, 07:46 PM
VNVeteran
Introduction to CONICS
Find the vertex,focus, and directrix of the parabola. Use a graphing utility to graph the parabola.

Y^2-4x-4=0
• Dec 6th 2009, 08:27 PM
Roam
Quote:

Originally Posted by VNVeteran
Find the vertex,focus, and directrix of the parabola. Use a graphing utility to graph the parabola.

Y^2-4x-4=0

I'm guessing this is what you need to do: comparing $y^2=4x+4$ with $y^2 = 4ax$ shows that a=1, so the focus is S(1,0). Find coordinates of one point on the curve to identify it. For example if x=2 then $y^2 =12$ and $y = \sqrt{12}$
And of course, the equation of the fixed line (directrix) is x=-a.