Question

What's the parabolic standard form of x^2 -12x - 48y -372 = 0

Answer 1

the standard form of a parabola is

\[y = ax^2 + bx + c\]

so rewrite you equation start by adding 48y to both sides of the equation.

then you well need to divide each term by 48 to get a single y

and subsequently, your equation in the form above.

Answer 2

try making perfect squares with x as the term x^2 is present

Answer 3

x^2 -12x - 48y -372 = 0
or
hint
x^2 -12x +36 = 48y +372 +36

Answer 4

wow I thought (x - h)^2 = 4a(y -k) was a vertex form

Answer 5

I suppose it depends on there you learn your maths

Answer 6

nope, you're right.

Answer 7

parabola is a geometric term, and as such should prolly ustilize the geometric standard definition :)

Answer 8

hmm, "(x - h)^2 = 4a(y -k)" standard focus form
"y = (x-h)^2 + k" standard vertex form

ax^2+b+c standard form

Answer 9

http://books.google.com/books?id=d2ZbnMdLeTQC&pg=PA642&dq=standard+form+parabola&hl=en&sa=X&ei=XYYKUrKeGOnx2AXI94CgAg&ved=0CDsQ6AEwAg#v=onepage&q=standard%20form%20parabola&f=false

Answer 10

$$
\tt From~x^2-12x-48y-372=0,~we~get~that~y=\frac{x^2}{48}-\frac{x}{4}-\frac{31}{4}\\
Assuming~Standard~form~is~(x-h)^2=4p(y-k) (^\star)\\
Where~(h,k)~is~vertex~and~p~is~focus.\\
We~have, with~y=ax^2+bx+c~,k=\frac{-b}{2a}~,h=\frac{4ac-b^2}{4a}~and~p=\frac{1}{4a}.\\
Substitute~h,~k~and~p~into~(^\star)~and~you're~done.\\
I ~get~ k=6,~h = -8.5~ and ~p = 12.
$$