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Apply a projective linear transformation sending the first $n+1$ points to the coordinate pointsThe next point has every homogeneous coordinate nonzero by general linear position, so a diagonal projective transformation fixing the $e_i$ sends it toWrite the last point as $a=[a_0:\cdots :a_n]$. Again general linear position implies that every $a_i$ is nonzero. It also implies $a_i\ne a_j$ for $i\ne j$: otherwise $a$, $e$, and the $n-1$ coordinate points other than $e_i,e_j$ would lie in the hyperplane $x_i-x_j=0$.

Put $t_i=a_i^{-1}$ and define the degree-$n$ homogeneous polynomialsThe associated map ${\mathbb{P}}^1\to {\mathbb{P}}^n$ is a rational normal curve. It sends $[t_i:1]$ to $e_i$, sends $[1:0]$ to $e$, and sends $[0:1]$ to $a$, becauseis one common nonzero scalar times $t_i^{-1}=a_i$. This proves existence.

For uniqueness, parametrize any rational normal curve through these points and choose a coordinate $[S:T]$ on its source ${\mathbb{P}}^1$ so that the preimages of $e$ and $a$ are respectively $[1:0]$ and $[0:1]$. If the preimage of $e_i$ is $[s_i:1]$, then its $i$th coordinate polynomial must have all the other $s_j$ as roots. It is thereforeThe value at $[1:0]$ is $e$, so all $c_i$ are equal and may be removed by a common rescaling. The value at $[0:1]$ is $a$, so $s_i^{-1}$ is proportional to $a_i$. Thus all $s_i$ are obtained from the $t_i$ by one common rescaling of the coordinate on ${\mathbb{P}}^1$. This merely reparametrizes the curve, proving uniqueness.

For $n=3$, choose the monomial basis $S^3,S^{2}T,S{T}^2,T^3$. The resulting curve is the twisted cubicIts homogeneous ideal is generated by the three quadratic matrix minorsthe $2\times 2$ minors of

The required general linear position condition is that every $n+1$ of the $n+2$ representing vectors are linearly independent. Such an ordered configuration is a projective frame.

Choose representatives $p_0,\dots ,p_n$ of the first $n+1$ projective points; they form a basis. Write a representative of the last point as $p_{n+1}={\sum }_i{a}_i{p}_i$. Every $a_i$ is nonzero: otherwise the last point together with the other $n$ basis points would fail the independence condition. Rescale the basis representatives to ${\tilde{p}}_i=a_i{p}_i$, so that $p_{n+1}={\sum }_i{\tilde{p}}_i$.

Do the same for the target configuration, obtaining representatives ${\tilde{q}}_i$ with $q_{n+1}={\sum }_i{\tilde{q}}_i$. The unique linear map sending the basis ${\tilde{p}}_i$ to ${\tilde{q}}_i$ is invertible and also sends $p_{n+1}$ to $q_{n+1}$. Its induced projective transformation sends every point to its prescribed partner. Thus any two ordered projective frames are projectively equivalent.

For the explicit basis in part (c), $q_0+t{q}_1$ has coefficients $(1+t,1-2t,t-2)$. ThusAt these values the conics factor, up to nonzero constants, asGeometrically these are exactly the three pairings of the four base points into two pairs, with a projective line joining each pair. No double line can contain four points in general linear position.

The numerical parameter values depend on the selected basis, but the three distinct singular members do not. Another pair of independent nonsingular basis conics gives an invertible projective change of pencil parameter. Its parameter infinity is nonsingular because the second basis form is nonsingular. Therefore all three singular members again occur at three distinct finite values of $t$. This proves the claimed count for any such basis.

Four projective points in general linear position in the complex projective plane impose four independent linear conditions on the six coefficients of a quadratic polynomial. To see independence, put them at the three coordinate points and $[1:1:1]$ using a projective frame. The remaining equations are $Axy+Bxz+Cyz=0$ with $A+B+C=0$, a two-dimensional vector space. Projectivizing its nonzero elements gives a pencil. It has nonsingular bases, for example the forms with coefficient triples $(1,1,-2)$ and $(1,-2,1)$, because the determinant of the symmetric coefficient matrix is $ABC/4$. Its three singular members are exactly the three pairs of lines obtained by partitioning the four base points into two pairs.