Show that the irreducible quartic projective curve is rational How can I show that the irreducible quartic curve $\Gamma=V_+((x^2-z^2)^2-y^2(2yz+3z^2))$ on $P^2(\mathbb{C})$ is rational by considering the family of conics through the double points $(1:0:1), (-1:0:1), (0:-1:1)$ and the simple point $(0:1:0)$.
I hope you can help. Thanks!
 A: Low-tech way, following your hint.
The family of conics passing through those points is 1-dimensional. This can be seen by counting dimensions (the space of conics is 5-dimensional, and each constraint subtracts one) or more explicitly by noting that the family is given by 
$$a(X^2-Z^2-YZ)+bZY=0$$
for $[a:b]\in \mathbb P^1$.
Conics are degree $2$, and your curve is degree $4$, so by Bezout's theorem, they intersect in $2\cdot 4=8$ points, counting multiplicity. The conic passes through 3 double points and 1 single point, so this accounts for 7 points, counting multiplicity. Then each choice of $[a:b]$ corresponds to another point on the curve (the remaining intersection point). This gives a map $\mathbb P^1\rightarrow \Gamma$, which I believe is dominant. (Check this!) Then $\Gamma$ is rational.
Higher-tech way. We invoke the degree-genus formula, noting the correction for ordinary singular points. We have 
$$\frac{(d-1)(d-2)}{2}=3,$$
and we have at least $3$ ordinary double points, so the geometric genus is at most $3-3=0$. But geometric genus must be non-negative, so $g=0$ and the curve is rational, as any genus $0$ curve is rational.
