SOLUTION: Supposedly this is a hyperbola: {{{ 16y^2-x^2+2x+64y+63=0 }}} rearranging the terms and completing the squares: {{{ 16(y^2+4y+4) - (x^2-2x+1) = -63+64-1 }}} Simplifying into "s

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Supposedly this is a hyperbola: {{{ 16y^2-x^2+2x+64y+63=0 }}} rearranging the terms and completing the squares: {{{ 16(y^2+4y+4) - (x^2-2x+1) = -63+64-1 }}} Simplifying into "s      Log On


   

Question 869735: Supposedly this is a hyperbola:
+16y%5E2-x%5E2%2B2x%2B64y%2B63=0+
rearranging the terms and completing the squares:
+16%28y%5E2%2B4y%2B4%29+-+%28x%5E2-2x%2B1%29+=+-63%2B64-1+
Simplifying into "standard form":
+16%28y%2B2%29%5E2+-+%28x-1%29%5E2+=+0+
But hyperbola in standard form should equal 1
What kind of shape is this equation?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Your steps seem correct. The shape for the resulting equation could be a degenerate hyperbola. It seems like a linear-like equation.

16%28y%2B2%29%5E2=%28x-1%29%5E2
Take square roots of both sides,
4%28y%2B2%29=0%2B-+%28x-1%29 ---- the zero is placed in the right hand member only to help with rendering; it is otherwise not needed.