Question 719601
Find the coordinates of the vertex,focus,ends of the latus rectum and the equation of the directrix.Draw the parabola of: 
x^2=12(y+7)
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This is an equation of a parabola that opens upwards:
Its basic form of equation:(x-h)^2=4p(y-k)
For given equation: x^2=12(y+7)
vertex: (0,-7)
axis of symmetry: x=0
4p=12
p=3
focus: (0,-4) (p-units above vertex on the axis of symmetry)
Ends of latus rectum(focal width):
plugs in y-coordinate of focus(-4) then solve for x
x^2=12(y+7)
x^2=12(-4+7)=12*3=36
x=±√36=±6
ends of latus rectum: (-6,-4) and (6,-4)

see graph below:

{{{ graph( 300, 300, -10, 10, -10, 10, (x^2-84)/12)}}}