Question 555476: y2-12x+2y=-37
identify the vertex, the focus, and the directrix of the graph of each equation. Then sketch the graph. Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! y2-12x+2y=-37
identify the vertex, the focus, and the directrix of the graph of each equation. Then sketch the graph.
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y2-12x+2y=-37
12x=y^2+2y+37
complete the square:
12x=(y^2+2y+1)+37-1
(y+1)^2=12x-36
(y+1)^2=12(x-3)
This is an equation of a parabola which opens rightwards of the standard form:
(y-k)^2=4p(x-h), with (h,k) being the (x,y) coordinates of the vertex.
For given equation:
vertex: (3,-1)
Axis of symmetry: y=-1
4p=12
p=3
focus=(6,-1) (p distance to the right of the vertex on the axis of symmetry
directrix: x=0 or y-axis (p distance to the left of the vertex on the axis of symmetry
see the graph below:
..
y=±(12x-36)^.5-1
(