Parabolas in standard form!
posted by Kate on .
Hello, I am having the worst time trying to solve these parabolas and putting them into xh = a(yk)^2 form. :(
There are two problems that i keep doing something wrong. could someone solve them so I can have a set up for the rest of my problems? :)
3y^2+6y+108x969=0
and
x^2+14x=44y+313=0
All help is greatly appreciated!

108x = 3y^2  6y + 969
108x = 3(y^2 + 2y + 1) + 972
x = 1/36 (y+1)^2 + 9
I'm not too sure how to do the second one. I'm really sorry. 
Can you solve this one?:
x^212x48y372=0
& I made a mistake one the second one!!
Its actually:
x^2+14x+44y+313=0 !! 
x^2+14x+44y+313=0
44y = x^2  14x  313
44y = (x^2 + 14x +4949  313
44y = ( (x+7)^2  49 )  313
44y = (x+7)^2 + 49  313
44y = (x+7)^2  264
44y + 264 = (x+7)^2
44(y + 6) = (x+7)^2
y + 6 = (1/44) (x+7)^2 , in the form that was requested.
x^212x48y372=0
x^2  12x  372 = 48y
x^2  12x +36 = 48y + 372 +36
(x6)^2 = 48y + 408
(x6)^2 = 48(y + 8.5)
y + 8.8 = (1/48) (x6)^2 
WOW!
Thank you so much! you made more sense than my teacher!
:D