Find an equation of the tangent plane to the surface z = 3y^2 − 2x^2 + x at the point P(2, −1, −3).

Question

Find an equation of the tangent plane to the surface z = 3y² − 2x² + x at the point P(2, −1, −3).

Answer 1

This problem is solved in three steps below.

As a preliminary check we should confirm that the point P(2, -1, -3)

does indeed lie on the given surface, which we can do by plugging the coordinates

of P into the given equation.

Step 1 is based on the fact that ∇F is perpendicular to any given surface ∇F = 0.

In out example

F(x, y, z) = 3y² - 2x² + x - z

∇F = (∂F/∂x, ∂F/∂y, ∂F/∂z) = (-4x + 1, 6y, -1)

At the point P(2, -1, -3)

∇F(2, -1, -3) = (-7, -6, -1)

Step 2 is based on the fact that any given vector (a, b, c) has perpendicular planes of the form

ax + by + cz = d, for any d.

In our example, the perpendicular planes are of the form

-7x - 6y - z = d

Step 3 computes d by finding that plane which passes through the given point P.

In our example,

d = -7×2 - 6×(-1) - (-3) = -5

Thus the solution is the plane

-7x - 6y - z = -5

Note that multiplying the equation by any constant, e.g., -1, gives us an equation of the same plane.

So another representation is

7x + 6y + z = 5