 Solution

Minimize or maximize $D(x)=x^{T}x$ subject to $x^{T}Ax=1$. A Lagrange multiplier $\Lambda$ givesThe constraint excludes $x=0$ and implies $Ax\ne 0$, so it is a regular constraint surface. Multiplying the stationarity equation by $x^T$ gives $D=\Lambda >0$. Thus the distance extrema on a real quadratic surface satisfyEvery constrained extremum lies in a positive-eigenvalue eigenspace, with squared distance $D=1/\lambda$. If $A$ is positive definite, the global minimum is $1/{\lambda }_{\max }$ and the global maximum is $1/{\lambda }_{\min }$. For an indefinite or singular form the surface need not be bounded; eigenvector directions give stationary candidates, which may be saddles and need not supply both global extrema. With a repeated positive eigenvalue, there can be an entire family of equally distant stationary points.