Ellipsoid: $$ x^2+\frac{y^2}{4}+\frac{z^2}{2}=1 $$ Sphere: $$ x^2+(y-1)^2+(z-d)^2=1 $$ For what values of $d$, there is a common tangent plane to both curves?
Part of my resolution:
Consider $f(x,y,z)=x^2+\frac{y^2}{4}+\frac{z^2}{2}\qquad$ and $g(x,y,z)=x^2+(y-1)^2+(z-d)^2$.
Then,
$f(x,y,z)=1$ and $g(x,y,z)=1$ are both level sets and a tangent plane (common) to them is given by, respectively: $$ \overrightarrow \bigtriangledown f(P_0) \cdot PP_0=0 $$ $$ \overrightarrow \bigtriangledown g(P_1) \cdot PP_1=0 $$ with $P_0=(x_0,y_0,z_0)$ and $P_1=(x_1,y_1,z_1)$ points of tangency.
I obtained the planes: $$ 2x_0x+\frac{y_0}{2} y+z_0z=2 $$ $$ x_1x+(y_1-1)y+(z_1-d)z=y_1+d(z_1-d) $$ Now, for a certain $\lambda \in \mathbb R$, we have: $$ x_1 = 2 \lambda x_0 $$ $$ y_1-1 = \lambda \frac{y_0}{2} $$ $$ z_1-d = \lambda z_0 $$ $$ y_1+d(z_1-d)=2 \lambda $$
I tried to solve this system but there are to many variables...
If $d\neq0$, we can easily see that the point $(0,2,d)$ on the sphere, in the plane of the biggest elliptical equator of the ellipsoid, is outside the ellipsoid. If $d=0$, then $(1/\surd2,\,1+1/\surd2,\,0)$ is clearly outside the ellipsoid. Hence, for all values of $d$, there is a point on the sphere outside the ellipsoid. The ellipsoid is too big to fit in the sphere; so some point of it must be outside the sphere. Hence there will always be parts of each surface outside the other surface, and so there will be a tangent plane common to them, whatever the value of $d$.