SOLUTION: Find the vertex, focus, and directrix of the parabola X^2-4X- 4Y +16=0. Solve and graph

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the vertex, focus, and directrix of the parabola X^2-4X- 4Y +16=0. Solve and graph       Log On


   

Question 1194362: Find the vertex, focus, and directrix of the parabola X^2-4X- 4Y +16=0. Solve and graph
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
X%5E2-4X-+4Y+%2B16=0
preferring all lower case,
x%5E2-4x-+4y+%2B16=0
x%5E2-4x=4y-16
x%5E2-4x%2B4=4y-16%2B4
%28x-2%29%5E2=4y-12
highlight_green%28%28x-2%29%5E2=4%28y-3%29%29
With this, if you compare to (x-h)^2=4p(y-k), then p=1, or the focus and directrix each is 1 unit from the vertex;
and
for your example,
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vertex (2,3)
parabola is with vertical symmetry axis with vertex as minimum;
focus (2,4)
directrix y=2
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