Volume without Multivariable Calculus

Question

The tangent plane to the ellipsoid at a point $(x_0, y_0, z_0)$ $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$ and the co-ordinate axis planes form a tetrahedron.

Only with the use of single variable calculus, find the volume of the tetrahedron.

I tried assuming a fixed value of $z$ and attempting to find the volume for unknown values of $x$ and $y$, however, the best I could do was find the area of the triangle bounded by $x=0, y=0, z = z_0$ and the tangent line on $z = z_0$

I also tried assuming a fixed ratio of $x:y$ but this essentially led to the same dead end as above.

Is it possible to find the volume of this tetrahedron given the restriction stated?

The use of vector based arguments is undesirable, but it is not restricted

Answer 1

If you have a tetrahedron $OABC$ then its volume is one sixth of that of the parallelepided including edges $OA$, $OB$, $OC$. Here the vertices of the tetrahedron are $(0,0,0)$, $(r,0,0)$, $(0,s,0)$ and $(0,0,t)$ where you will need to find $r$, $s$ and $t$, so the volume is $|rst|/6$.

Answer 2

let $u = \frac xa, v = \frac yb, w = \frac zc$

$u^2 + v^2 + w^2 = 1$

The plane tangent to the point $(u_0, v_0,w_0)$ has an equation of $u_0u+v_0v + w_0w = 1$ and intersects the coordinate axes at.

$(\frac {1}{u_0}, 0,0), (\frac {1}{v_0}, 0,0), (\frac {1}{v_0}, 0,0)\\ (\frac {a}{x_0}, 0,0), (\frac {b}{y_0}, 0,0), (\frac {c}{z_0}, 0,0)$

And translating back to $x,y,z$ space

$(\frac {a^2}{x_0}, 0,0), (\frac {b^2}{y_0}, 0,0), (\frac {c^2}{z_0}, 0,0)$

Together with $(0,0,0)$

The volume of this tetrahedron is $\frac 16 \frac {a^2b^2c^2}{x_0y_0x_0}$

I guess that isn't single variable calculus either, it is just algebra.