# Equation for stationary values of $px^2+qy^2+rz^2$ given sphere and plane constraints

Consider stationary points of the function $V=px^2+qy^2+rz^2$ subject to the constraints $x^2+y^2+z^2=1$ and $lx+my+nz=0$, where $l,m,n$ not all zero and $p,q,r$ not all equal. How can we show that the stationary values of $V$ satisfy $\frac{l^2}{V-p} + \frac{m^2}{V-q} + \frac{n^2}{V-r} = 0$?

I've shown that the stationary point $(x,y,z)$ must satisfy $l(q-r)yz + m(r-p)xz + n(p-q)xy = 0$, but can't see how to convert this into the required equation. It would be sufficient to show that $(q-r)yz(px^2+qy^2+rz^2-p)=l$ (and similar conditions with $p,q,r$ and $l,m,n$ and $x,y,z$ cycled), but is this true? What about if we add $lx+my+nz$ to the equation I've got?

Thanks for any help with this!

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By "stationary" you mean, that all $(x,y,z)$ share the same valuer of $V$? – draks ... Apr 1 '12 at 17:54
I mean $V$ takes its maximum or minimum value. – Harry Macpherson Apr 1 '12 at 18:20

Rearrange the equation I already got to $px(ny-mz)+qy(lz-nx)+rz(mx-ly)=0$, which shows that $\begin{pmatrix} px\\qy\\rz \end{pmatrix}$ is perpendicular to $\begin{pmatrix} x\\y\\z \end{pmatrix}\times\begin{pmatrix} l\\m\\n \end{pmatrix}$. So $\begin{pmatrix} px\\qy\\rz \end{pmatrix}=V\begin{pmatrix} x\\y\\z \end{pmatrix}+c\begin{pmatrix} l\\m\\n \end{pmatrix}$, which means $\begin{pmatrix} V-p\\V-q\\V-r \end{pmatrix}=-c\begin{pmatrix} l/x\\m/y\\n/z \end{pmatrix}$. Then take reciprocals of components, sum them, and use the given constraint $lx+my+nz=0$.