I am reading Rick Miranda's "Linear systems of plane curves". A cubic hypersurface in $\mathbb{P}^{4}$ that passes through $7$ points in general position with multiplicity $2$ is not expected to exist. Nevertheless, it exists, and I am trying to understand Miranda's example.
There exists a rational normal curve $X$ that passes through $7$ points in general position in $\mathbb{P}^{4}$. Let $S(X)$ be its secant variety. $S(X)$ is supposed to be the variety we are looking for.
The secant variety of an irreducible curve is three-dimensional unless the curve is contained in a plane. A rational normal curve in $\mathbb{P}^{n}$ spans $\mathbb{P}^{n}$, so it is not contain in a plane. According to this, dim$(S(X))=3$.
Now, why is the degree of $S(X)$ equal to $3$? I think it suffices to show that any $l\in\mathbb{G}(1,4)-S(X)$ satisfies $\#X\cap l=3$, but I am not able to prove it.
Now, let's suppose that $S(X)$ passes through any of the seven points $p$ with multiplicity $1$. Then dim $T_{p}S(X)=3$. Since $X$ spans $\mathbb{P}^{4}$, there are secant lines to $X$ that contain $p$ in $4$ independent directions. I am not able to understand why this fact contradicts dim $T_{p}S(X)=3$. Is $\{l\in\mathbb{G}(1,4):\text{l is secant to $X$ and $p\in l$}\}\subseteq T_{p}S(X)$?
Any help would be appreciated.
From [BO]
[BO] Maria Chiara Brambilla and Giorgio Ottaviani, On the Alexander-Hirschowitz theorem, J. Pure Appl. Algebra 212 (2008)
[CH] C. Ciliberto, A. Hirschowitz, Hypercubiques de P4 avec sept points singuliers génériques. C. R. Acad. Sci. Paris Sér. I Math. 313 (1991), no. 3, 135-137