SOLUTION: The vertex,focus and directrix of the parabola y^2+4x-16y+8=0

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Question 972479: The vertex,focus and directrix of the parabola y^2+4x-16y+8=0
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
A reference model for a horizontal parabola is 4px=y%5E2 and allowing for translations of location, 4p%28x-h%29=%28y-k%29%5E2. These forms come from the derivation for the equation of a parabola. The meaning of p is the distance between the vertex (for standard position) and either the focus or the directrix.

Start from the general given equation, complete the square, and put into STANDARD FORM.

y%5E2-16y=-4x-8
y%5E2-16y%2B64=-4x-8%2B64, the constant, 64 needed for completing the square for y.
%28y-8%29%5E2=-4x%2B56
-%28y-8%29%5E2=4%28x-14%29
-4%28x-14%29=%28y-8%29%5E2
-------still not exactly standard form, but vertex is (14,8), then corresponding coefficients gives 4p=-4 and then p=-1.
The vertex occurs as a maximum for x value so the parabola opens to the left, and therefore the focus is 1 unit to the left of the vertex.
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Focus is (13,8);
Directrix is x=15.
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