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

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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
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#1
+79734
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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
+149
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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

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