SOLUTION: 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)

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)
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
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)
...
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:

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