Question 552789
the focus of a parabola is (5,5) and directrix is y= -3 . 
write an equation for the parabola and then draw the graph.
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Given position of directrix shows this is a parabola which opens upwards.
Its standard form: (x-h)^2=4p(y-k), with (h,k) being the (x,y) coordinates of the vertex.
Axis of symmetry: x=5
p=half the distance between focus and directrix on the axis of symmetry=8/2=4
y-coordinate of vertex= midpoint between focus and directrix on the axis of symmetry=(5-3)/2=1
vertex: (5,1)
Equation:
(x-5)^2=16(y-1)
or
y=(1/16)(x-5)^2+1
see graph below:
{{{ graph( 300, 300, -10, 10, -10, 10,(1/16)(x-5)^2+1) }}}