SOLUTION: I was given the equation: x^2+4x-6y=-10 it says write the standard equation for the parabola. state the vertex, focus, and directrix. i started off with adding 6y to both sides

Question 604104: I was given the equation: x^2+4x-6y=-10
it says write the standard equation for the parabola. state the vertex, focus, and directrix.
i started off with adding 6y to both sides. then i added 4 to both sides.
now i have
(x+2)^2=6y-6
i am stuck. please help
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
6y = (x + 2)^2 + 6

y = (1/6)(x + 2)^2 + 1 ___ this is the vertex form, with vertex at (-2,1)

the vertex is midway between the focus and the directrix; with the vertex and focus located on the axis of symmetry,
and the directrix perpendicular to it

the distance (p) from the the vertex to the focus (or the directrix) is │1 / 4a│
___ in this case, 6/4 or 3/2

you can think of the focus as being "inside" the curvature of the parabola, while the directrix is "outside"

with this upward opening parabola (a is positive), the focus is above the vertex and the directrix is below it

so the focus is (-2,5/2); and the directrix is y = -1/2
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