This should be straightforward, but honestly I forgot even the names Google for...
I've got a surface description in this form (what is it called?):
$$z=\dfrac{cr^2}{1+\sqrt{1-(1+k)c^2r^2}},$$
where $r=\sqrt{x^2+y^2}$. This form is standard in optics. How can it be converted to the quadric form? I need it to program ray-tracing, and I haven't used the quadric form since university.
$$z=\dfrac{cr^2}{1+\sqrt{1-(1+k)c^2r^2}},$$
rationalise the denominator
$$ z = \frac{cr^2\left(1+\sqrt{1-(1+k)c^2r^2}\right)}{\left(1-\sqrt{1-(1+k)c^2r^2}\right)\left(1+\sqrt{1-(1+k)c^2r^2}\right)} \\ =\frac{cr^2\left(1+\sqrt{1-(1+k)c^2r^2}\right)}{(1+k)c^2r^2} \tag{1} $$
then re-write Eq.(1) as $$ cr^2\sqrt{1-(1+k)c^2r^2} = z(1+k)c^2r^2 - cr^2 $$ I think you can take it from here right?
$\textbf{Update:}$
since the OP has disappeared and no complete answer (or at least I think its an answer) $$ \sqrt{1-(1+k)c^2r^2} = z(1+k)c - 1 $$ thus $$ 1 - (1+k)c^2r^2 = \left(z(1+k)c - 1 \right)^2 = c^2(1+k)^2z^2 -2c(1+k)z+1\\ \implies (1+k)c^2r^2 + c^2(1+k)^2z^2 -2c(1+k)z=Ar^2 + Bz^2+Cz = 0\ $$