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Let $p_1,...,p_8\in\mathbb{P}^{3}$ be points in linear general position. Then there exists a unique elliptic curve $C$ of degree $4$ passing through $p_1,...,p_8$. I am interested in what happens for nodal rational curves of degree $4$.

Now, let us suppose we have seven points $p_1,...,p_7\in\mathbb{P}^{3}$ in linear general position.

How many irreducible, rational curves of degree $4$ pass through $p_2,...,p_7$ and have a singular point of multiplicity $2$ at $p_1$ ?

Such a curve can be constructed as a complete intersection of two quadric surfaces which are tangent at $p_1$, or as a projection of a rational normal quartic curve $C$ in $\mathbb{P}^{4}$ from a point lying on a secant line or an a tangent line of $C$.

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Zero. Indeed, if the intersection $Q_1 \cap Q_2$ of two quadrics is singular at $p_1$, then there is a quadric $Q$ in the pencil generated by $Q_1$ and $Q_2$ which is singular at $p_1$. On the other hand, if a quadric cone with vertex at $p_1$ passes through $p_2,\dots,p_7$, then the images of $p_2,\dots,p_7$ in $\mathbb P^2$, obtained by the projection from $p_1$, all lie on a conic. But a general 6-tuple of points in $\mathbb P^2$ does not lie on a conic.

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  • $\begingroup$ This would imply that the following situation is impossible: There exist a point $p\in\mathbb{P}^{4}$ and a rational normal curve $C\subset\mathbb{P}^{4}$ such that $\left\langle p,p_1\right\rangle$ is secant to $C$ and $\left\langle p,p_i\right\rangle$ intersects $C$ for any $i = 2,...,7$. Is this clear to you? $\endgroup$ – F_L Mar 8 '14 at 16:52
  • $\begingroup$ $p_1,\cdots,p_7$ are points in $\mathbb P^3$. If you want to lift them to $\mathbb P^4$, you don't get to change the point $p$. Otherwise by varying $p$, you could get different arrangements of $7$ points in $\mathbb P^3$. So you probably want to fix $p,p_1,\cdots,p_7$. Then this is generically impossible for the same reason - projection to $\mathbb P^2$. Alternately, you can view this as a dimension-counting argument, I believe. $\endgroup$ – Will Sawin Mar 8 '14 at 20:05
  • $\begingroup$ Just two comments: 1) your argument is nice and but you don't need in fact to consider any cones or quadrics, just project directly from $p_1$ and observe that the image of the quartic curve is a conic. 2) In the question it was written "in general linear position", but what you probably mean is "in general position", because there are indeed $7$ pts in general linear position where the projection by one of the pts is a plane an the image of the $6$ others are on a conic. $\endgroup$ – Jérémy Blanc Mar 8 '14 at 22:06
  • $\begingroup$ @Jeremy Blanc: you are right of course, it is much simpler to project the curve and do not care about the quadrics. $\endgroup$ – Sasha Mar 9 '14 at 15:57

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