SOLUTION: Find the vertex, focus, and directrix of the parabola given by the equation x^2 + 4x + 4y + 16 + 0

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Question 857856: Find the vertex, focus, and directrix of the parabola given by the equation x^2 + 4x + 4y + 16 + 0
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Something is missing. An red%28equa%29tion must have an red%28equa%29l sign.
Maybe you meant to type x%5E2+%2B+4x+%2B+4y+%2B+16+=+0 and you got a + sign instead of the = sign.

x%5E2+%2B+4x+%2B+4y+%2B+16+=+0<-->x%5E2%2B4x=-4y-16<-->x%5E2%2B4x%2B4=-4y-16%2B4%29<-->%28x%2B2%29%5E2=-4y-12<-->%28x%2B2%29%5E2=-4%28y%2B3%29<-->y%2B3=-%281%2F4%29%28x%2B2%29%5E2<-->y=-%281%2F4%29%28x%2B2%29%5E2-3
is the equation of a parabola with the vertical axis of symmetry
x%2B2=0<-->x=-2 ,
and the vertex at (-2,-3).
The focal distance is 1 ,
and the focus and the rest of the parabola is below (-2,-3).