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olya-2409 [2.1K]
3 years ago
11

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.
Mathematics
1 answer:
stealth61 [152]3 years ago
6 0

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).

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