Science:Math Exam Resources/Courses/MATH200/December 2011/Question 04/Solution 1

If the sphere with radius in question $\displaystyle R$ is enclosed by the ellipse, and the radius is maximal, then the surface of the ellipse and the sphere just touch in one point.

The tangent planes to the surface $\displaystyle 2(x+1)^{2}+y^{2}+2(z-1)^{2}=8$ and the sphere with radius $\displaystyle R$ are parallel at the points where the surface and the sphere touch. Hence the normal vectors to the tangent planes are also parallel.

Regarding the surface equation and the sphere equation as level sets of functions $f,g:\mathbb {R} ^{3}\rightarrow \mathbb {R}$, then the normal vectors to these tangent planes are given by the gradients of the functions $\displaystyle f$ and $\displaystyle g$.

The sphere equation with the origin as center and radius $\displaystyle R$ is $\displaystyle x^{2}+y^{2}+z^{2}=R^{2}$. Then $\displaystyle f(x,y,z)=2(x+1)^{2}+y^{2}+2(z-1)^{2}$ and $\displaystyle g(x,y,z)=x^{2}+y^{2}+z^{2}$. We are looking for $\displaystyle x,y,z,\lambda$ such that the gradients are parallel,

${\begin{aligned}\nabla f&=\lambda \nabla g\\{\begin{pmatrix}4(x+1)\\2y\\4(z-1)\end{pmatrix}}&=\lambda {\begin{pmatrix}2x\\2y\\2z\end{pmatrix}}\end{aligned}}$

The second line yields that $\displaystyle \lambda =1$ or $\displaystyle y=0$.

• $\displaystyle \lambda =1$ leads to $\displaystyle x=-2,z=2$ and using the surface equation we obtain the points $\displaystyle p_{1}=(-2,2,2),\ \ p_{2}=(-2,-2,2)$
• $\displaystyle y=0$ leads to $\displaystyle x=-z$ and using the surface equation we obtain the points $\displaystyle p_{3}=({\sqrt {2}}-1,0,-{\sqrt {2}}+1),\ \ p_{4}=(-{\sqrt {2}}-1,0,{\sqrt {2}}+1)$.

These four points are candidates for where the ellipse and the sphere touch. Since we are searching for the radius $\displaystyle R$ such that the sqhere is enclosed in the ellipse, we need the point $\displaystyle p_{i}$ with the shortest distance to the origin.

${\begin{aligned}||p_{1}||&={\sqrt {12}}\\||p_{2}||&={\sqrt {12}}\\||p_{3}||&={\sqrt {6-4{\sqrt {2}}}}\\||p_{4}||&={\sqrt {6+4{\sqrt {2}}}}\end{aligned}}$

Since $\displaystyle {\sqrt {6-4{\sqrt {2}}}}$ is the smallest of these numbers, this is the radius $\displaystyle R$ of the sqhere.