Show that the sum of the squares of the reciprocals of any three diameters of an ellipsoid which are mutually at right angle is constant.
I tried assuming $(x_i,y_i, z_i), i=1,2,3$ as three points on ellipsoid, say $P,Q,R$ respectively. Then $(x_1/OP,y_1/OP, z_1/OP)$, $(x_2/OQ,y_2/OQ, z_2/OQ)$, and $(x_3/OR,y_3/OR, z_3/OR)$ represent direction cosines of three mutually perpendicular directions.
We need to effectively prove, find $1/OP^2+ 1/OQ^2+1/OR^2 = 1/a^2+1/b^2+1/c^2$. But I am not able to proceed. Any help greatly appreciated!
I just realized the solution. Take $(x_1, y_1,z_1)= OP(l_1,m_1,n_1)$ where OP is length of semidiameter represented by point P. Similarly represent other points. And $(l_i,m_i,n_i)$ are direction cosines of three mutually perpendicular directions.
Use that $x_1^2/a^2+y_1^2/b^2+z_1^2/c^2 =1$ and above transformation to get $1/OP^2 = l_1^2/a^2+m_1^2/b^2+n_1^2/c^2$ and similar equations for other semidiameters. Adding these proves the statement.