Question

Let II be the tangent plane to the graph of f(x, y) = 8 – 2x^2 – 3y^2 at the point (1, 2,-6). Let S, x² + y^2 + z = 4 be another surface. Find the point on S which has tangent plane parallel to I.

Answer 1

Let $F(x,y,z)=f(x,y)-z$. The tangent plane to $f(x,y)$ at (1, 2, -6) has equation

$\nabla F(1,2,-6)\cdot(x-1,y-2,z+6)=0$

We have

$\nabla F(x,y,z)=(-4x,-6y,-1)\implies\nabla F(1,2,-6)=(-4,-12,-1)$

Then the tangent plane has equation

$(-4,-12,-1)\cdot(x-1,y-2,z+6)=0\implies -4(x-1)-12(y-2)-(z+6)=0\implies 4x+12y+z=22$

Let $g(x,y)=4-x^2-y^2$, and $G(x,y,z)=g(x,y)-z$. The tangent plane to $S$ at a point $(a,b,c)$ is

$\nabla G(a,b,c)\cdot(x-a,y-b,z-c)=0$

We have

$\nabla G(x,y,z)=(-2x,-2y,-1)\implies \nabla G(a,b,c)=(-2a,-2b,-1)$

so that this plane has equation

$(-2a,-2b,-1)\cdot(x-a,y-b,z-c)=0\implies2ax+2by+z=2a^2+2b^2+c$

In order for this plane to be parallel to the previous plane, we need to have

$\begin{cases}2a=4\\2b=12\end{cases}\implies a=2,b=6\implies g(a,b)=c=-36$

so the point we're looking for is (2, 6, -36).