SOLUTION: what is the focus diretrix and and axis of symmetry for -x^2=48y i think the focus is (0,12) but im not sure

Question 1001325: what is the focus diretrix and and axis of symmetry for -x^2=48y
i think the focus is (0,12) but im not sure
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
48y= -x^2
y= -(1/48)x^2. This is a parabola that opens downward, is wide due to the small coefficient of x, and is symmetrical around the y-axis. The focus will be negative, the directrix positive.
(y-0)^2= (-1/48)(x-0)^2
4p(y-k)=(x-h)^2
y^2=(-1/48)(x-0)^2
-48y^2=(x-0)^2
4p=-48
p=-12
The focus is 12 units below the vertex, which is at (0,0). That would be at (0,-12)
The axis of symmetry is the y-axis, or x=0.
The directrix is 12 units above the vertex or y=12

Check a point. At x=6, y=-36/48 or -3/4. The distance of the point from the directrix is 12.75 units.
The distance of that point (6,-0.75) from the focus is sqrt (6^2+11.25^2)=sqrt(162.5625)=12.75 units

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