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.
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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