The equation I want to graph is: $$4x^2 -3y^2 + 8x +12y-4 =0$$
My textbook says that the above equation has no graph.
Yet, I rearranged it to the following:
$$\frac{(y-2)^2}{(2/\sqrt3)^2} - \frac{(x+1)^2}{1^2} = 1^2$$
Therefore, I thought it was a hyperbola. I know I've made a mistake in the rearrangement, as the two equations produce different graphs in Desmos. However, both equations produce hyperbolas, which contradicts my textbook.
Where am I going wrong? Thanks.
If the book indeed says "no graph", it is wrong.
Because when plugging $(10,0)$, you get a positive value, and when plugging $(0,10)$, a negative one. Hence there are two regions of opposite signs and there must be a continuous curve in-between.
But my bet is that the book meant something different.
Note that when $x,y$ are both large, the equation virtually reduces to
$$2x=\pm\sqrt3y$$ which gives the directions of the asymptotes of the… hyperbola.