SOLUTION: How do I identify the focus, directrix and axis of symmetry when y= -1/12x^2?

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Question 1011138: How do I identify the focus, directrix and axis of symmetry when y= -1/12x^2?
Found 2 solutions by MathLover1, It is costly:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
+y=+-%281%2F12%29x%5E2+
+%281%2F%28-1%2F12%29%29y=+x%5E2+
+x%5E2+=%2812%2F-1%29y
x%5E2+=-12y
we can see that the vertex will be at (h,k)=(0,0)
The standard form of a parabola (in this case upside down parabola) with a vertical axis of symmetry is
%28x-h%29%5E2=-4p%28y-k%29, where p is the directed distance between the vertex and focus (and also between the vertex and directrix)
since h=0 and k=0, we have
%28x-0%29%5E2=-4p%28y-0%29
x%5E2=4p%2Ay since you have x%5E2+=-12y, then
4p=-12
p=-12%2F4
p=-3
focus | (0,p)=(0, -3)
p is the directed distance between the vertex and directrix, directrix will be 3 units above x-axis :
y+=+3






Answer by It is costly(175) About Me  (Show Source):