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I can interpolate a hyperbolic paraboloid given 4 points in space ($pmba_1, pmba_2, pmbb_1, pmbb_2$) with the following parametric equation for a ruled surface:

$$ pmbr(u, v) = (1-v)((1-u)pmba_1+upmba_2) + v((1-u)pmbb_1+ upmbb_2) $$

Can I obtain the equation of the surface in cartesian coordinates?

You could painstakingly eliminate the parameters $u$ and $v$ (or have a computer do it for you), but if you’re performing the calculations yourself I think it might involve less work to transform the canonical hyperbolic paraboloid (“hypar”) $X^2-Y^2=Z$, represented by the matrix $$Q = beginbmatrix1&0&0&0\0&-1&0&0\0&0&0&-frac12\0&0&-frac12&0endbmatrix.$$ Using your parameterization, this hypar can be generated by the points $$mathbf a_1' = (-1,0,1)^T \ mathbf a_2' = (0,-1,-1)^T \ mathbf b_1' = (0,1,-1)^T \ mathbf b_2' = (1,0,1)^T,$$ so if $M$ maps these points to the ones that define the desired hypar, its associated matrix is $M^-TQM^-1$. The transformation matrix $M$ is easily constructed from the point pairs, but we really need $$M^-1 = beginbmatrixmathbf a_1' & mathbf a_2' & mathbf b_1' & mathbf b_2' \ 1&1&1&1endbmatrix beginbmatrixmathbf a_1 & mathbf a_2 & mathbf b_1 & mathbf b_2 \ 1&1&1&1endbmatrix^-1 = A'A^-1.$$ We can precompute part of $M^-TQM^-1=A^-TA'^TQA'A^-1$ to simplify the eventual computation a bit: $$Q'=A'^TQA' = beginbmatrix0&0&0&-2\0&0&2&0\0&2&0&0\-2&0&0&0endbmatrix$$ so that the matrix of the hypar is given by the product $A^-TQ'A^-1$. Since we’re working in homogeneous coordinates, we can multiply this matrix by $1/2$ (or any other nonzero scalar) for convenience without changing the resulting quadric.

For example, let $$mathbf a_1 = (-2,-1,2)^T \ mathbf a_2 = (-1,1,0)^T \ mathbf b_1 = (2,-2,1)^T \ mathbf b_2 = (3,0,2)^T.$$ We then have $$A^-1 = beginbmatrix-2&-1&2&3\-1&1&-2&0\2&0&1&2\1&1&1&1endbmatrix^-1 = beginbmatrix-frac527&-frac227&frac13&-frac19 \ -frac127&frac527&-frac13&frac79 \ frac227&-frac1027&-frac13&frac49\ frac427&frac727&frac13&-frac19 endbmatrix.$$ Multiplying this by $27$ to eliminate the fractions and using $frac12Q'$ yields $$beginbmatrix36&63&0&27\63&-72&0&-135\0&0&0&-243\27&-135&-243&486endbmatrix.$$ Pulling out another factor of $9$, this corresponds to the equation $$4x^2+14xy-8y^2+6x-30y-54z+54=0.$$ (Another factor of $2$ can be pulled out of this equation, if desired.) Plotting both the parametric patch and this hypar in Geogebra produces the following:

You could instead start from the canonical equation $XY=Z$, with associated matrix $$beginbmatrix0&frac12&0&0\frac12&0&0&0\0&0&0&-frac12\0&0&-frac12&0endbmatrix$$ and go through a similar calculation after selecting a suitable set of points to generate this hypar. If you precompute $Q'$ as above, you should end up with a scalar multiple of the previous matrix.