SOLUTION: the focus and directrix of a parabola are given. the focua is (3,5) and the directrix is y=1 . write the equation of the parabola and then draw the graph.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: the focus and directrix of a parabola are given. the focua is (3,5) and the directrix is y=1 . write the equation of the parabola and then draw the graph.      Log On


   



Question 552780: the focus and directrix of a parabola are given. the focua is (3,5) and the directrix is y=1 . write the equation of the parabola and then draw the graph.
Answer by lwsshak3(11628) About Me  (Show Source):
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the focus and directrix of a parabola are given. the focua is (3,5) and the directrix is y=1 . write the equation of the parabola and then draw the graph.
Standard form of equation for parabola: (x-h)^2=4p(y-k), (h,k) being the (x,y) coordinates of the vertex.
For given equation:
Parabola opens upwards
axis of symmetry: x=3
p=half the distance between directrix and focus on the axis of symmetry=4/2=2
y-coordinate of vertex=3
vertex: (3,3)
Equation of parabola:
(x-3)^2=8(y-3)
or
y=(1/8)(x-3)^2+3
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