Find the point of tangency between a plane and an ellipsoid

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?

• Thanks for your help everybody! – 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 $$k = \pm 8$$ NOTE

At tangency we have

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

then

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

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

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

• Thanks a bunch, this really helped! – 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 $$4x^2+y^2+2z^2=16$$

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

$$\implies x=\pm1$$

$$(x,y,z)\equiv\pm(1,2,2)$$

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