# Tangent points of two surfaces

## Homework Statement

Determine if the surfaces $$x^3 + 6xy^2 + 2z^3 = 48$$ and $$xyz = 4\sqrt{2}$$ have any points of tangency, and if so, find those points.

## The Attempt at a Solution

I'm mostly wondering if anyone can find any problems with my approach.

I assume that the geometry of the problem implies that a point of tangency should be a point of intersection between the two surfaces, and that the tangent planes of the two surfaces should be equal in this point. Or, equivalently, the two (non-null) normal vectors of the surfaces are parallell in the point of intersection.

Furthermore, a normal vector of the first surface is

$$\nabla (x^3 + 6xy^2 + 2z^3 - 48) = \left(3x^2+6y^2, 12xy, 6z^2\right)$$

and a normal vector of the second surface is

$$\nabla (xyz - 4\sqrt{2}) = \left(yz,xz,xy)$$.

Since these vectors must be parallell, we get

$$\begin{cases} 3x^2 + 6y^2 = \lambda yz\\ 12xy = \lambda xz\\ 6z^2 = \lambda xy \end{cases}$$.

for some non-zero $$\lambda \in {R}$$.

After a bit of work, it can be found that $$\lambda = \pm 6 \sqrt{2} ($$if $$x \ne 0)$$. Inserting this into the equation system above, it is found that $$x=z, y=\pm\frac{z}{\sqrt{2}}$$ (A). Thus, any point of intersection between the two surfaces that fulfils (A) will be a point of tangency between the surfaces.

Using the restriction (A) on the equations of the two surfaces, it is found that the point $$P = (2,\sqrt{2},2)$$ is the only point that fulfils both equations. Thus, $$P$$ is the only point of tangency.