If I have found the canonical form to be $x^2/2^2-y^2/3^2-1=0$. Do I simply sketch this as I would a usual $(x,y)$ graph? As when I typed this into a graphing calculator I got 4 lines (2 straight and 2 curved). So I am not too sure if its right.
Also when working the original equation and getting it into canonical form I am asked to state the translation and orthogonal transformation? What do these exactly refer to.
$5x^2 + 12xy − 22x − 12y − 19 = 0$ was the original equation.
Solve for $y$ and get two solutions (one positive, one negative):
The "canonical" form of such an unrotated conic is:
$$\frac{(x-a)^2}{b^2} + \frac{(y-c)^2}{d^2} = 0,$$
placing the center at $(a,c)$ with axes of length $b$ and $d$.