The equation of a parabola is given. y=−1/12x^2−2x−1

What are the coordinates of the focus of the parabola?

Kakarot_15
May 3, 2017

#1**+1 **

y=−1/12x^2−2x−1 factor as

y = (-1/12) [ x^2 + 24x + 12 ] complete the square on x

y = ( -1/12) [ x^2 + 24x + 144 + 12 - 144]

y = (-1/12) [ ( x + 12)^2 - 132 ]

y = (-1/12) (x + 12)^2 + 11

(y - 11) = (-1/12) (x + 12)^2

The vertex is ( -12, 11) and this parabola opens downward

And 4p = -1/12 → p = - 1 / 48

So....the coordinates of the focus are ( -12 - 1/48, 11) = ( - 577/48, 11 )

CPhill
May 3, 2017

#2**0 **

I don't want to question you as you are at the top of the list but you have me confused for the final answer

Kakarot_15
May 4, 2017