SOLUTION: could someone help me with this problem? the focus of the parabola 4y^2+12x-12y+39=0 is________? how do i find this? thanks for your help..it was a problem presented in clas

Question 74666: could someone help me with this problem?
the focus of the parabola 4y^2+12x-12y+39=0 is________?
how do i find this? thanks for your help..it was a problem presented in class.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
the focus of the parabola 4y^2+12x-12y+39=0 is________?
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You need to find the vertex and the value of "p".
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Rewrite as 4y^2-12y =-12x-39
Complete the square on the left side:
4(y^2-3y+(3/2)^2) = -12x-39+4(3/2)^2
4(y-(3/2))^2 = -12x-39+9
(y-(3/2))^2 = -3x-30/4
(y-(3/2))^2 = -3(x+(10/4))
This is the form of a parabola opening to the left.
So, the vertex is at (-10/4,(3/2))
And 4p=-3 so p=-3/4
Therefore the focus is at (-10/4-(3/4),3/2) or (-13/4,3/2)
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Cheers,
Stan H.
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