Fresnel's wave surface

Remarks

Away from the coordinate spheres $r = v_j$, the equation is equivalent to Fresnel's form $\sum_j v_j^2 x_j^2 / (r^2 - v_j^2) = 0$. Take the point at distance $r$ from the origin in a direction $\mathbf{t}$. It lies on the surface exactly when $r$ is the ray speed of one of the two modes whose energy flows along $\mathbf{t}$. So the surface is the wavefront at time $1$ of a point source at the origin.

Example: in a uniaxial medium with $v_x = v_y = v_o$ and $v_z = v_e$, the left side factors as $$\left(r^2 - v_o^2\right)\left(v_o^2 (x^2 + y^2) + v_e^2 z^2 - v_o^2 v_e^2\right).$$ So the surface is the sphere of radius $v_o$, the ordinary wave, together with the spheroid $(x^2 + y^2)/v_e^2 + z^2/v_o^2 = 1$, the extraordinary wave. The two sheets touch at the two points $(0, 0, \pm v_o)$ on the optic axis.

Non-example: in an isotropic medium with all $v_j = v$, the left side becomes $v^2 (r^2 - v^2)^2$, so the equation is $(r^2 - v^2)^2 = 0$. This is a doubly counted sphere, and the medium is not birefringent.

In a biaxial medium with distinct $v_j$, the two sheets meet in four conical points that lie on the two optic ray axes. The wave surface is dual to the surface of wave normals built from the index ellipsoid.