I want to know the number of smooth conics that tangent 4 general lines and pass through 1 general point. Indeed, the number is 2.
But from Bezout's theorem we get $1\cdot2^4=16$ on $\mathbb{P}^5$. The set of conics tangent line has degree 2 and theset of pass through point has degree 1 in $\mathbb{P}^5$. We think $(Q_1, Q_2, Q_3, Q_4)$ as a section of $\cal{O}_{\mathbb{P}^4}$(2)$^4$. $\quad h^i =0 \ (i \geq 2).\quad h \in H^2(\mathbb{P}^1),\ H \in H^2(\mathbb{P}^4).\quad H=2h.$
$ \begin{eqnarray} c(N_{\mathbb{P}^1/\mathbb{P}^4})&=&c(T\mathbb{P}^4)|_{{\mathbb{P}^1}}/c(T{\mathbb{P}^1})\\ &=&((1+H)^5)|_{\mathbb{P}^1}/(1+h)^2 \\ &=& (1+2h)^5(1+h)^{-2}\\ &=&1+8h. \end{eqnarray}$
Then
$\begin{eqnarray} c(B)&=&((1+2H)^4|_{\mathbb{P}^1})(1+8h)^{-1} \\ &=&(1+4h)^4(1+8h)^{-1}\\ &=&1+8h. \end{eqnarray}$
$c_1(B)=8h$. Therefore, the excess contribution is 8.
THIS IS WRONG, but I don't think that there is mistake in calculation. I checked [Fulton] but there isn't detail only described "$\cdots$14 in $\mathbb{P}^4$".
I want 14. Please some clue or answer.