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

Question 869735: Supposedly this is a hyperbola:

rearranging the terms and completing the squares:

Simplifying into "standard form":

But hyperbola in standard form should equal 1
What kind of shape is this equation?
Answer by josgarithmetic(39617) (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.

Take square roots of both sides,
---- the zero is placed in the right hand member only to help with rendering; it is otherwise not needed.
RELATED QUESTIONS
graph the hyperbola with the following equation.... (answered by Alan3354)

Find the vertices, foci, and center of the hyperbola. Graph the equation. 4y^2 - x^2 -... (answered by lwsshak3)

What would the standard graphing form of these following equations be? I don't know if my (answered by Fombitz)

how can i find the focus/foci and vertex/vertices x^2-4y^2-2x+16y=20 this is how... (answered by solver91311,lwsshak3)

6. find the center, vertices, and foci of the ellipse with the following equation: (answered by lynnlo)

Completing the square 1. X^4+64y^4 2. X^4+y^2+25 3. X^4-11x^2y^2+y^4 Please do... (answered by dkppathak)

Completing the square 1. X^4+64y^4 2. X^4+y^2+25 3. X^4-11x^2y^2+y^4 Please do... (answered by KMST)

x^2-2x-4y^2+16y=11 (answered by ewatrrr)

1. 5 over 2x plus x over 2 2. 4y2(this is y squared)... (answered by tutorcecilia)