Why is the volume of this tetrahedron 1/6 times the product of side lengths? Given an ellipsoid $$x^2/a^2 + y^2/b^2 + z^2/c^2 = 1$$
And bounding a volume that is a tetrahedron with 4 planes, the $x=y=z=0$ planes and the plane tangent to the ellipsoid at some point $(x_0, y_0, z_0)$, why is the volume of this tetrahedron 
$$ \frac {1}{6} \frac {a^2b^2c^2}{x_0y_0z_0} $$
Can someone please offer some intuition on how this formula is derived?  Wikipedia, Wolfram and other problems that I have looked up regarding volumes of tetrahedron have not helped much.  
Is $\large \frac {a^2}{x_0}$ the "length" of a side of the tetrahedron?
Thanks,
 A: First let's find the volume of an easier tetrahedron, namely a tetrahedron found from the planes $x=y=z=0$ and $Ax+By+Cz=D$. It is nothing but $\frac{1}{6}\frac{D}{A}\frac{D}{B}\frac{D}{C}$, where $\frac{D}{A}$ etc are the lengths of the sides.
Now back to your example we only need to find the tangent plane at $x_0,y_0,z_0$ on the ellipsoid. Since the ellipsoid is given as a contour equation:
$$
f(x,y,z) = \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1
$$
The gradient of $f$ is normal to the ellipsiod at any point, in particular $\mathbf{n}=\nabla f(x_0,y_0,z_0) = \frac{2x_0}{a^2}\hat{x}+\frac{2y_0}{b^2}\hat{y}+\frac{2z_0}{c^2}\hat{z}$ is the normal vector for the tangent plane at $\mathbf{v}_0=(x_0,y_0,z_0)$ on the ellipsoid. The plane is given simply as $\mathbf{n}\cdot (\mathbf{v}-\mathbf{v}_0)=0$ or
$$
0=\frac{x_0(x-x_0)}{a^2}+\frac{y_0(y-y_0)}{b^2}+\frac{z_0(z-z_0)}{c^2}
$$
which can be rewritten as (using the ellipsoid equation)
$$
\left(\frac{x_0}{a^2}\right)x+\left(\frac{y_0}{b^2}\right)y+\left(\frac{z_0}{c^2}\right)z = 1
$$
Now by what we found in the beginning the volume is simply
$$V=\frac{a^2b^2c^2}{6x_0y_0z_0}$$
and yes $\frac{a^2}{x_0}$ is the sidelength.
