Question 35839
Hi! I'm working on some Conic Section problems, and do ok when they are centered at zero. My problem is when their vertex has shifted and I am asked to graph a problem with a given equation or if I'm given the foci/directrix and am required to write the equation. If you could clear this up for me that would be wonderful!

Write the standard equation:
y^2-4x+4=0
STANDARD FORM OF PARABOLA IS
(Y-K)^2=4A(X-H),WHERE (H,K) IS VERTEX,FOCUS IS (H+A,K),DIRECTRIX IS X-H+A=0,
AND AXIS IS Y-K=0.HENCE WE GET HERE AS STANDARD FORM 
(Y-0)^2=4(X-1)....H=1...K=0...A=1...
VERTEX = (1,0)
FOCUS IS (1+1,0)=(2,0)
DIRECTRIX IS X-1+1=0....OR....X=0
AXIS IS Y-0=0...Y=0
---------------------------------------------------------------
Find the standard form of the equation of the parabola with focus (8,-2) and directrix x=4
FOCUS .....H+A=8......I................K=-2
DIRECTRIX=X-H+A=0...OR....X=H-A=4............II
EQN.I+EQN.II GIVES
H+A+H-A=8+4=12
2H=12
H=6
A=8-H=2
HENCE EQN.IS
(Y-K)^2=4A(X-H)
(Y+2)^2=4*2*(X-6)=8(X-6)

Thanks so much for taking time to look at my question!
~Troubled Conic Student :)