So, it is given that -

The tangent plane to the ellipsoid $4x^2 + y^2 + 2z^2 = 16$ is $2x + y + 2z = k$.

I’m trying to find k, and the point of tangency between those two.

What I did - Assumed that the normal to the plane is parallel to the gradient of the ellipsoid, so I get $(8x, 2y, 4z) = (2, 1, 2)$. But then I get point $(x, y, z)$ which isn’t on the ellipsoid.

Why is my assumption wrong?

  • $\begingroup$ Thanks for your help everybody! $\endgroup$ – John Dec 15 '18 at 21:12

Substituting the plane equation into the ellipsoid we have

$$ 4 x^2 + (k - 2 x - 2 z)^2 + 2 z^2 = 16 $$

Solving for $x$ we have

$$ x = \frac{1}{4} \left(k-2 z\pm\sqrt{4 k z-8 z^2-k^2+32}\right) $$

but at tangency we need

$$ 4 k z-8 z^2-k^2+32 = 0 $$

solving now for $z$

$$ z = \frac{1}{4} \left(k\pm \sqrt{64-k^2}\right) $$

but at tangency $64-k^2 = 0$ then $k = \pm 8$

enter image description here


At tangency we have

$$ (8x,2y,4z)=\lambda(2,1,2) $$


$$ x = \frac{\lambda}{4}\\ y = \frac{\lambda}{2}\\ z = \frac{\lambda}{2} $$

now after substitution into the ellipsoid equation we have the $\lambda$ value. Now with the tangency point, after substitution into the plane equation we get the $k$ value

  • $\begingroup$ @amd Please. See attached note. $\endgroup$ – Cesareo Dec 15 '18 at 20:00
  • $\begingroup$ @amd Yes. I agree with you. Thanks. $\endgroup$ – Cesareo Dec 15 '18 at 20:06

You’ve set the two vectors equal to each other, which is too strong. If they’re parallel, then all you can say for sure is that one is a scalar multiple of the other, i.e., you need $(8x,2y,4z)=\lambda(2,1,2)$ for some $\lambda\ne0$ instead. Since you’re working in three dimensions, you can avoid introducing an extraneous variable by stating the condition as $(8x,2y,4z)\times(2,1,2)=0$ instead. This will give you three linear equations, two of which are independent.

  • $\begingroup$ Thanks a bunch, this really helped! $\endgroup$ – John Dec 15 '18 at 21:12

The vector $(8x,2y,4z)$ is parallel to the vector $(2,1,2)$, not equal to it.

$8x=4z\implies z=2x, 2\times2y=8x\implies y=2x$

Substitute for $y,z$ in $4x^2+y^2+2z^2=16$

$\implies 4x^2+4x^2+8x^2=16x^2=16$

$\implies x=\pm1$


Depending on the value of $k$, you will have to select one solution.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.