SOLUTION: y^2 - 16x^2 + 14y + 64x - 31 = 0 i moved 31 to the right and tried to perfect square it, in sure sure if im doing it correctly. this is a hyperbola right? thanks in advance :)

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Question 598932: y^2 - 16x^2 + 14y + 64x - 31 = 0
i moved 31 to the right and tried to perfect square it, in sure sure if im doing it correctly. this is a hyperbola right? thanks in advance :)
Answer by KMST(5347) About Me  (Show Source):
You can put this solution on YOUR website!
It is a hyperbola with axes parallel to the x- and y-axes.
y%5E2+-+16x%5E2+%2B+14y+%2B+64x+-+31+=+0 ---> %28y%5E2+%2B+14y%29+-+16%28x%5E2+-4x%29+=+31
would be a good start.
Then you would complete the squares:
%28y%5E2%2B14y%2B49%29+-+16%28x%5E2-4x%2B4%29+=+31%2B49-16%2A4 --> %28y%2B7%29%5E2-16%28x-2%29%5E2=16
Dividing everything by 16, we get
%28y%2B7%29%5E2%2F16-16%28x-2%29%5E2%2F16=1 --> %28y%2B7%29%5E2%2F4%5E2-%28x-2%29%5E2%2F1%5E2=1
So the hyperbola is centered at (2,-7).
The transverse axis is vertical x=2.
When x=2 --> %28y%2B7%29%5E2=4%5E2
The hyperbola crosses the transverse axis at points (2,5) and (2,-19)
with y=4-7=-3 and y=-4-7=-11
The asymptotes cross at center (2,-7), and have the equations
y%2B7=4%28x-2%29 and y%2B7=-4%28x-2%29
that derive from %28y%2B7%29%5E2%2F12%5E2=%28x-2%29%5E2%2F3%5E2
Here is half of that hyperbola, with the asymptotes:
graph%28300%2C300%2C-3%2C7%2C-8%2C12%2C4sqrt%281%2B1%28x-2%29%5E2%29-7%2C-4x%2B1%2C4x-15%29