Question

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

Answer 1

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]