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Question 542145: Please help me with this problem:
How do I put the following in standard form??
y^2-6y+16x+25=0
I know it is a parabola because only one variable is squared and I know that it opens vertically because the y is what is squared. I know the standard form of the parabola is (x-h)=(1/4p)((y-2)^2) but i dont know how to get it there.
this is what i tried:
y^2-6y+16x+25=0
(y^2-6y+___ )+16x=-25+__ I am completing the square
(y^2-6y+9)+16x=-25+9
(y-3)^2+16x=-16
well now i am stuck becasue if i didvide by sixteen i lose the x coefficient. I have been sitting here stairing at the problem for like 30min hoping the answer would come to me but no such luck!!
If you can explain this problem to me I would highly appreciate it!! Thank You!!
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! You're doing great. I'll start where you left off.
(y-3)^2+16x=-16
(y-3)^2=-16-16x
(y-3)^2=-16x-16
(y-3)^2=-(16x+16)
-(y-3)^2=16x+16
16x+16=-(y-3)^2
16(x+1)=-(y-3)^2
(x+1)=-(1/16)(y-3)^2
(x+1)=-(1/(4*4))(y-3)^2
(x-(-1))=(1/(4*(-4)))(y-3)^2
The equation is now in the form (x-h)=(1/4p)(y-k)^2 where h = -1, p = -4, and k = 3
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