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

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

We have

`\\abla F(x,y,z)=(-4x,-6y,-1)\implies\\abla 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

`\\abla G(a,b,c)\cdot(x-a,y-b,z-c)=0`

We have

`\\abla G(x,y,z)=(-2x,-2y,-1)\implies \\abla 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).