Restricting a function on the sphere to a one-parameter family of arcs Let's consider the following constrained optimization problem:

Given the quadratic form $x^TAx$ defined using the matrix $A$ below, find the maximum and minimum values subject to the constraint $\|x\| = 1$ along with their location on the sphere.
$$\begin{bmatrix} 2 & 1 & 0\\ 1 & 2 & 0\\ 0 & 0 & 1\end{bmatrix}$$

Since the maximum and minimum eigenvalues are $3$ and $1$, with eigenvectors
$$A = \frac{1}{\sqrt{2}}\begin{bmatrix} 1\\ 1\\ 0 \end{bmatrix}; \quad \frac{1}{\sqrt{2}}\begin{bmatrix} -1\\ 1\\ 0 \end{bmatrix}; \quad \begin{bmatrix} 0\\ 0\\ 1\end{bmatrix}$$
this gives us our answer. However, I was curious about the behavior of this quadratic form on the rest of the sphere. How does it behave outside of the intersection of the sphere and these eigenspaces? Well, we can restrict ourselves to the small arc on the $xy$-plane that runs from $\frac{1}{\sqrt{2}}(-1,1,0)$ to $\frac{1}{\sqrt{2}}(1,1,0)$. If we do this, our quadratic form is
$$2+2x_1x_2$$
and has the following behavior:

where we start at the blue line, the minimum at $\frac{1}{\sqrt{2}}(-1,1,0)$ and move to the red line, the maximum at $\frac{1}{\sqrt{2}}(1,1,0)$. To explore this further, I wanted to move this arc along the unit circle on the $1$-eigenspace plane -- fix the endpoint at $\frac{1}{\sqrt{2}}(1,1,0)$, and slide the beginning point from $\frac{1}{\sqrt{2}}(-1,1,0)$ to $\frac{1}{\sqrt{2}}(0,0,1)$.
This should yield a one-parameter family of functions that I should be able to throw into Desmos and see how the family of functions changes as this parameter varies from $0$ to $\pi/2$.
How exactly can I cook up this parameterized family?

I tried using a rotation matrix $R_t$ to compute $(R_tx)^TA(R_tx)$ and restrict this quadratic form to my original arc, but I ended up getting
$$(2+2\sin^2(t))x_1\sqrt{1-x_1^2} + (1+\cos^2(t))$$
which doesn't always attain a minimum at $1$, which it should if I did everything correctly.
 A: Any ellipse can be parameterized as $\mathbf r(t)=\mathbf c+\mathbf u\cos t+\mathbf v\sin t$, where $\mathbf c$ is its center and the vectors $\mathbf u$ and $\mathbf v$ correspond to a pair of conjugate half-axes. In the special case of a circle, $\mathbf u$ and $\mathbf v$ are orthogonal and have lengths equal to the circle’s radius.  
To parameterize an arc of a great circle on the unit sphere, you need a normal $\mathbf n$ of the circle’s plane. If the end points $\mathbf p_1$ and $\mathbf p_2$ of the arc aren’t antipodes, you can compute this as either $\mathbf p_1\times\mathbf p_2$ or $\mathbf p_2\times\mathbf p_1$, depending on whether you want the minor or major arc, respectively, with those endpoints. For the parameterization of the arc, take $\mathbf u=\mathbf p_1$ and for $\mathbf v$ use $\mathbf n\times\mathbf u$, normalized. The center is, of course, the origin. With this parameterization, $\mathbf p_1=\mathbf r(0)$ and $\mathbf p_2$ is attained when $t$ is equal to the angle subtended by the arc. You can of course scale the parameter if you want to normalize the domain to $[0,\pi/2]$. For the quadratic form restricted to the arc we then have $$\begin{align} \mathbf r(t)^TA\mathbf r(t) &= (\mathbf u\cos t+\mathbf v\sin t)^TA(\mathbf u\cos t+\mathbf v\sin t) \\
&= \mathbf u^TA\mathbf u\cos^2t+\mathbf v^TA\mathbf v\sin^2 t+\mathbf v^TA\mathbf u\sin2t. \end{align}$$ 
For your example, since you’ve chosen endpoints that are an angle of $\pi/2$ apart, the endpoints are $\mathbf u$ and $\mathbf v$, so the arc has the parameterization $$\mathbf r(t)=\frac1{\sqrt2}(\sin t-\cos t,\sin t+\cos t,0)^T$$ and its composition with the quadratic form simplifies to $2-\cos2t$.  
Geometrically, the value of the quadratic form at a unit vector $\mathbf u$ is equal to the reciprocal square of the distance from the origin of the intersection of the ray from the origin with direction $\mathbf u$ and the quadric surface $\mathbf x^TA\mathbf x=1$.  
A bit more generally, you can also investigate the behavior on circular arcs that aren’t on great circles using this same parameterization. Now $\mathbf u$ will be the orthogonal rejection of $\mathbf p_1$ from $\mathbf n$ and the length of $\mathbf v$ must be adjusted so that $\lVert\mathbf v\rVert = \lVert\mathbf u\rVert$.
