SOLUTION: The equation of a parabola is given. y=−1/12x^2−2x−1 What are the coordinates of the focus of the parabola?

First get into standard form, which is



Then the vertex is (h,k), the focus is (h,k+p),
and the equation of the directrix is y=k-p
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Instead of doing your homework for you, I'll
do one exactly like it.  Then you can use it 
as a model to do yours step by step



Clear of fractions by multiplying through by 20



Get the x terms on the left side
and all the other terms on the right



Complete the square on the left:

1. Multiply the coefficient of x, which is 40, by 1/2, getting 20
2. Square what you got, 202=400
3. Add that, 400, to both sides of the equation



Factor the left side as (x+20)(x+20) and write it as (x+20)2
Combine the -200+400 on the right as +200



Factor out the coefficient of y, which is -20 on the right,
remembering that when you factor out a negative, you get
a sign change inside the parentheses:



Compare that to



where the vertex is (h,k), the focus is (h,k+p),
and the equation of the directrix is y=k-p

So h = -20, 4p = -20, k=10.

And since 4p =-20, p = -5.

So the vertex is (-20,10), the focus is (-20,10-5) or (-20,5)

(-20,5)  <-- focus

and the equation of the directrix is y=10-(-5) or y = 15 



Now you can do yours exactly the same way step by step.

Edwin