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

Y^2-4x-4=0

Printable View

- Dec 6th 2009, 07:46 PMVNVeteranIntroduction 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 PMRoam

I'm guessing this is what you need to do: comparing $\displaystyle y^2=4x+4$ with $\displaystyle y^2 = 4ax$ 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 $\displaystyle y^2 =12$ and $\displaystyle y = \sqrt{12}$

And of course, the equation of the fixed line (directrix) is x=-a.