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.

Let . The tangent plane to at (1, 2, -6) has equation

We have

Then the tangent plane has equation

Let , and . The tangent plane to at a point is

We have

so that this plane has equation

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

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

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

may i see the picture? it would a lot, xox