- #1

anosh_88

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## Homework Statement

Let P be the tangent to the graph of g(x,y) = 8-2x^2-3y^2 at the point (1, 2, -6). Let f(x,y) = 4-x^2-y^2. Find the point on the graph of f which has tangent plane parallel to P.

## Homework Equations

g(x,y) = 8-2x^2-3y^2 at (1, 2, -6)

f(x,y) = 4-x^2-y^2

## The Attempt at a Solution

I found P, which is P = 4x+12y-34

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