Homework Help: Parabolas in standard form!

Posted by Kate on Thursday, May 24, 2012 at 9:06pm.

Hello, I am having the worst time trying to solve these parabolas and putting them into x-h = a(y-k)^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+108x-969=0

and

x^2+14x=44y+313=0

All help is greatly appreciated!

Parabolas in standard form! - Neil, Thursday, May 24, 2012 at 9:12pm
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.
Parabolas in standard form! - Kate, Thursday, May 24, 2012 at 9:21pm
Can you solve this one?:

x^2-12x-48y-372=0

& I made a mistake one the second one!!
Its actually:

x^2+14x+44y+313=0 !!
Parabolas in standard form! - Reiny, Thursday, May 24, 2012 at 9:42pm
x^2+14x+44y+313=0
44y = -x^2 - 14x - 313
44y = -(x^2 + 14x +49-49 - 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^2-12x-48y-372=0
x^2 - 12x - 372 = 48y
x^2 - 12x +36 = 48y + 372 +36
(x-6)^2 = 48y + 408
(x-6)^2 = 48(y + 8.5)

y + 8.8 = (1/48) (x-6)^2
Parabolas in standard form! - Kate, Thursday, May 24, 2012 at 9:48pm
WOW!

Thank you so much! you made more sense than my teacher!

:D

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