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 :)

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) (Show Source): You can put this solution on YOUR website!
It is a hyperbola with axes parallel to the x- and y-axes.
--->
would be a good start.
Then you would complete the squares:
-->
Dividing everything by 16, we get
-->
So the hyperbola is centered at (2,-7).
The transverse axis is vertical .
When -->
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
and
that derive from
Here is half of that hyperbola, with the asymptotes:

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