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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

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    $\begingroup$ You don’t really need calculus at all. A tangent plane intersects a quadric in exactly one point. That should be enough to work out the equation of the tangent plane to any point on the ellipsoid. $\endgroup$ – amd Apr 14 '18 at 5:02
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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.

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  • $\begingroup$ finding the tangent plane equation without calculus is non-trivial $\endgroup$ – Jack Lam Apr 19 '18 at 9:52
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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 $|rst|/6$.

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  • $\begingroup$ Yes, but how would you find those values without the aid of multi-variable calculus? This would have been better left as a comment. $\endgroup$ – Jack Lam Apr 14 '18 at 4:46

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