Another commonly used normalization is to set F=1. If we use the same notations as in the last subsection, the problem becomes to minimize the following function:
where is the sixth element of vector , i.e., .
Indeed, we seek for a least-squares solution to under the constraint . The equation can be rewritten as
where is the matrix formed by the first (n-1) columns of , is the last column of and is the vector . The problem can now be solved by the technique described in sect:Ax=b.
In the following, we present another technique for solving this kind of problems, i.e.,
based on eigen analysis, where we consider a general formulation, that is is a matrix, is a m-vector, and is the last element of vector . The function to minimize is
As in the last subsection, the symmetric matrix can be decomposed as in (5), i.e., . Now if we normalize each eigenvalue and eigenvector by the last element of the eigenvector, i.e.,
where is the last element of the eigenvector , then the last element of the new eigenvector is equal to one. We now have
where and . The original problem (7) becomes:
Find such that is minimized with
subject to .
After some simple algebra, we have
The problem now becomes to minimize the following unconstrained function:
where is the Lagrange multiplier. Setting the derivatives of with respect to through and yields:
The unique solution to the above equations is given by
where . The solution to the problem (7) is given by
Note that this normalization (F=1) has singularities for all conics going through the origin. That is, this method cannot fit such conics because they require to set F=0. This might suggest that the other normalizations are superior to the F=1 normalization with respect to singularities. However, as shown in , the singularity problem can be overcome by shifting the data so that they are centered on the origin, and better results by setting F=1 has been obtained than by setting A+C=1.