What is the shortest distance between the plane Ax + By + Cz + 1 = 0 and the ellipsoid x^{2}/a^{2} + y^{2}/b^{2} + z^{2}/c^{2} = 1. You may find it convenient to use the notation h = (A^{2} + B^{2} + C^{2})^{-1/2}, m = (a^{2}A^{2} + b^{2}B^{2} + c^{2}C^{2})^{1/2}. What is the algebraic condition for the plane not to intersect the ellipsoid?

**Solution**

The tangent plane to the ellipsoid at (X, Y, Z) is Xx/a^{2} + Yy/b^{2} + Zz/c^{2} = 1. It is parallel to Ax + By + Cz + 1 = 0 iff X/a^{2} = kA, Y/b^{2} = kB, Z/c^{2} = kC for some k. But 1 = X^{2}/a^{2} + Y^{2}/b^{2} + Z^{2}/c^{2} = k^{2}(a^{2}A^{2} + b^{2}B^{2} + c^{2}C^{2}) = k^{2}m^{2}, so k = ±1/m. There are two values corresponding to two parallel tangent planes (one on either side of the ellipse). The equation of the tangent plane is k(Ax + By + Cz) = 1.

The distance of the origin from the plane Ax + By + Cz + 1 = 0 is 1/(A^{2} + B^{2} + C^{2})^{1/2} = h. The distance of the origin from the tangent plane k(Ax + By + Cz) = 1 is h/|k| = hm. So if m ≥ 1, the plane Ax + By + Cz + 1 = 0 lies between the two tangent planes and hence intersects the ellipse. So in this case the minimum distance is zero. If m < 1, then the distance between the plane Ax + By + Cz + 1 = 0 and the nearer tangent plane is h(1 - m) and that is the required shortest distance.

© John Scholes

jscholes@kalva.demon.co.uk

5 Mar 2002