what are the coordinates of the focus of the conic section shown below? y^2+16y-4x+4=0

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what are the coordinates of the focus of the conic section shown below? y^2+16y-4x+4=0

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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start by completing 2 squares x's and y's
did you skip a term with x^2?

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can you go through the steps please? im really confused
|dw:1441343632292:dw|
Got it, or more explanation?
|dw:1441343759380:dw|
the answer was actually (-14,-8)
but thank you anyways to both of you!! :)
Really?? Your welcome!
y^2 + 16 = 4x - 4 completing the square (y + 8)^2 - 64 = 4x - 4 (y + 8)^2 = 4x + 60= 4(x + 15) compare with standard form (y - k)^2 = 4a(x - h) vertex is at (-15,-8) 4a = 4 so a = 1 focus is at (-15+1, -8) = (-14,-8)

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