It's given in my text book that the tangent at the origin can found out by equating to zero the lowest degree terms in $x$ and $y$. Therefore, by manipulating-
$yx^2-16y-2x^2+8=0$, the lowest degrees terms are $-16y-2x^2$, therefore- $x=\pm \sqrt{-8y}$, which is imaginary. Thus, we have a conjugate.
But the graph of $y=\frac{2x^2-8}{x^2-16}$ doesn't prove the findings.
Am I doing something wrong?