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delta59
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Aug2905, 07:27 AM (EST) 

1. "RE: Focus of parabola."
In response to message #0

The vertex of any parabola lies on the axis of symmetry and is equidistant from the focus and the directrice. In your case the vertex is V(3,1). Assume focus F has coordinates F(3,1+t) so directrice has equation y = 1 – t. Find any point on the parabola, for example A(2,0). Then AF^^2 = (23)^^2 + (0(1+t))^^2 = t^^2 – 2t + 2. And the distance from A to the directrice is t+1. So (t+1)^^2 = t^^2 – 2t + 2 .This gives t = 1/4.


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