Tangent points of two surfaces

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Homework Statement



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



Homework Equations





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

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

and a normal vector of the second surface is

[tex]\nabla (xyz - 4\sqrt{2}) = \left(yz,xz,xy)[/tex].

Since these vectors must be parallell, we get

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

for some non-zero [tex]\lambda \in {R}[/tex].

After a bit of work, it can be found that [tex]\lambda = \pm 6 \sqrt{2} ([/tex]if [tex] x \ne 0)[/tex]. Inserting this into the equation system above, it is found that [tex]x=z, y=\pm\frac{z}{\sqrt{2}}[/tex] (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 [tex]P = (2,\sqrt{2},2)[/tex] is the only point that fulfils both equations. Thus, [tex]P[/tex] is the only point of tangency.
 

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