I have a function for a plane: $x + z = d$ and a function for an ellipsoid $x^2 + y^2 + 2z^2 = 1$

The problem is to determine if there exists a number d, such that the plane is tangential to the ellipsoid.

I know that the normal for the plane is (1,0,1) and the gradient of the ellipsoid, which gives the normal, is (x, y, 2z). If the plane and the ellipsoid have the same normal vector at a point then they are tangential at that point.

Can I say that they both have the same normal vector at $x = 1, y = 0, z = \frac{1}{2}$ ?

There are some tips provided with the question. It first says that the function for the ellipsoid is "locally a function f of two variables". I'm not really sure about how I'm supposed to interpret this. Should I rearrange the function as $z = f(x,y)$?

It also says that I should determine the function for the tangent plane for the ellipsoid at the point $(a, b, f(a,b))$ and compare this with the plane x + z = d. This should give me an equation system containing a, b and d, but I haven't been able to get that equation system since I always end up with the dependent variable $f(a,b)$ in the equation which complicates things.

  • $\begingroup$ Note that the statement in your title is not quite the same as the one in the question. Going with the one in the question, "The problem is to determine if there exists a number $d$ such that...", this needs no calculation at all. Clearly the origin is inside the ellipsoid, so if $d=0$ then the plane $x+z=d$ cuts the ellipsoid in a curve. Now gradually change $d$: this moves the plane parallel to its original position; at some point the intersection with the ellipsoid will be just about to disappear, and at this point the plane will be tangent to the ellipsoid. $\endgroup$ – David May 1 '14 at 13:52
  • $\begingroup$ The point is that "the same normal" is not equivalent to "parallel normals". If you want to solve starting from implicit equations, try $<x,y,2z>=k<1,0,1>$ . $\endgroup$ – Tony Piccolo May 3 '14 at 16:42

Solution starting from explicit equations:

If you want to follow the tips, solve (two times) the system $$\begin {cases} f_x(a,b)=-1 \\ \\f_y(a,b)=0 \end {cases}$$ to find $$d=a+f(a,b)$$ with $$f(a,b)=\pm \sqrt {\frac {1-a^2-b^2}2}$$You'll find two solutions.


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