Consider this parametrization $$\phi:\mathbb{P}^1\longrightarrow\mathbb{P}^3$$
$$(t_0:t_1)\longmapsto (t_0^3: t_0^2t_1:t_0t_1^2:t_1^3)$$
Let $\mathcal{C}$ be the image of $\phi$. I've proved that $\mathcal{C}=V(F,G,H)$, where $V$ means zeri locus and $F(X_0,X_1,X_2,X_3)=X_0X_3-X_1X_2, G=X_0X_2-X_1^2,H=X_1X_3-X_2^2$. If we call $\mathcal{Q}_F,\mathcal{Q}_G,\mathcal{Q}_H$ the quadrics with equations $F,G,H$ respectively, we have
$$\mathcal{C}=\mathcal{Q}_F\cap\mathcal{Q}_G\cap\mathcal{Q}_H$$ Now I want to prove that each pair of previous quadrics intersects in $\mathcal{C}$ union a projective line. For instance, let $L$ be the following line $$L:X_1=X_2=0$$ Then $L\subseteq \mathcal{Q}_G\cap\mathcal{Q}_H$. My professor said:
$\operatorname{deg}(\mathcal{Q}_F\cap\mathcal{Q}_G)=4$
$\deg{\mathcal{C}}=3$
By Bezout Theorem, $\mathcal{Q}_G\cap\mathcal{Q}_H=\mathcal{C}\cup L$. My questions are:
1) Why $\operatorname{deg}(\mathcal{Q}_F\cap\mathcal{Q}_G)=4$?
2) Why $\deg{\mathcal{C}}=3$? To find the degree of a curve, I suppose I should know the equation of that curve, but I have only a parametric representation of $\mathcal{C}$, how can I deduce that degree is 3?
3) Bezout theorem which I listened about concerns the number of intersections of two plane curves of given degrees.....is this another different Bezout theorem? How can I apply it to quadrics??