Sequences and series

Question

Let P be the tangent plane to the surface z = x² + 2y² at the point Q= (-1, -1,3). Which of the following points also lies in P?

(a). (-1,0,1)

(b) (-1,0,-1)

(c). (0,-1,-1)

(d). (0,-1,2)

the answer is (b), but I am confused about how I do the working.

Answer 1

For x² + 2y² = z, or x² + 2y² - z = 0, the gradient of the function is (2x, 4y, -1), so the tangent plane is

(2x0)x0 + (4y0)y0 + (-1)z0 = c

As (xo, y0, z0) = (-1, -1, 3), we have

2(-1)² + 4(-1)² + (-1)3 = c

2 + 4 - 3 = c

c = 3

-2x -4z - z = 3, or

2x + 4y + z = -3

If we plug in (b) = (-1, 0, -1), we get

2(-1) + 4(0) + 1(-1) = -2 + 0 - 1 = -3; this is on the plane.

Plug in the other 3 points, and you will see that none of them gives a value of 3.