What is the focus and vertex of the parabola described by #3x^2+1x+2y+7=0 #?

1 Answer
Binayaka C.
Jun 11, 2016

Vertex is at # =(-1/6, -83/24)# Focus is at # (-1/6,-87/24)#

Explanation:

#2y=-3x^2-x-7 or y = -3/2 x^2-x/2-7/2=-3/2(x^2+x/3+1/36)+1/24-7/2 = -3/2(x+1/6)^2-83/24# Vertex is at # =(-1/6, -83/24)# The parabola opens down as co -efficient of #x^2# is negative. distance between vertex and focus is #1/|4a|=1/(4*3/2)=1/6# Hence the focus is at # -1/6,(-83/24-1/6) or (-1/6,-87/24)# graph{-3/2x^2-x/2-7/2 [-20, 20, -10, 10]}[Ans]

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