### 1st Putnam 1938

Problem B6

What is the shortest distance between the plane Ax + By + Cz + 1 = 0 and the ellipsoid x2/a2 + y2/b2 + z2/c2 = 1. You may find it convenient to use the notation h = (A2 + B2 + C2)-1/2, m = (a2A2 + b2B2 + c2C2)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/a2 + Yy/b2 + Zz/c2 = 1. It is parallel to Ax + By + Cz + 1 = 0 iff X/a2 = kA, Y/b2 = kB, Z/c2 = kC for some k. But 1 = X2/a2 + Y2/b2 + Z2/c2 = k2(a2A2 + b2B2 + c2C2) = k2m2, 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/(A2 + B2 + C2)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.