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Find the equation of the tangent plane to the surface z= x²+4xy-6y² at the point (4, -4, -144).

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Final answer:

The equation of the tangent plane to the surface z = x² + 4xy - 6y² at the point (4, -4, -144) is z = -64y - 400.

Step-by-step explanation:

To find the equation of the tangent plane to the surface z = x² + 4xy - 6y² at the point (4, -4, -144), we first need to find the partial derivatives of z with respect to x and y. The partial derivative with respect to x is 2x + 4y, and the partial derivative with respect to y is 4x - 12y. At the point (4, -4), these derivatives are 2(4) + 4(-4) = 0 and 4(4) - 12(-4) = 64, respectively.

The equation of the tangent plane at a point (a, b, c) of a surface z = f(x, y) is given by z - c = fx(a, b)(x - a) + fy(a, b)(y - b), where fx and fy are the partial derivatives at the point (a, b). In this case, the equation is z + 144 = 0(x - 4) + 64(y + 4).

Simplifying, we find the equation of the tangent plane to be z = -64y - 400.

User Malcolm McCaffery
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