Question

trace the curve y^2 = x^2 + y - 6

Answer 1

Algebraic curve

Answer 2

\[y^2-y-x^2=-6\]
\[(y-1/2)^2-x^2=1/4-6=-23/4\]
\[x^2-(y-\frac{1}{2})^2=\frac{23}{4}\]
looks like a hyperbolar

Answer 3

@jonask How about a circle with center at (0,1/2) and radius of sqrt(23/4)
(x-a)^2+(y-b)^2 =r^2 standard equation of a circle

Answer 4

these a minus between the brackets so its a problem (x-a)^2-(y-b)^2 =r^2

Answer 5

ohh Crap! sorry I didn't see your solution properly :(