Show that the ellipsoid 3x2 +2y2 +z2 = 9 and the sphere x2 +y2 +z2 −8x−6y −8z +24 = 0are tangent to each other at the point (1, 1, 2)

To show that the ellipsoid and the sphere are tangent at the point (1,1,2), we need to show two things:

1. The point (1,1,2) lies on both surfaces.
2. The gradients of the two surfaces at the point (1,1,2) are proportional.

Step 1: Check if the point (1,1,2) lies on both surfaces.

For the ellipsoid 3x^2 + 2y^2 + z^2 = 9, substitute x=1, y=1, z=2:

3(1)^2 + 2(1)^2 + (2)^2 = 3 + 2 + 4 = 9

So, the point (1,1,2) lies on the ellipsoid.

For the sphere x^2 + y^2 + z^2 - 8x - 6y - 8z + 24 = 0, substitute x=1, y=1, z=2:

(1)^2 + (1)^2 + (2)^2 - 8(1) - 6(1) - 8(2) + 24 = 1 + 1 + 4 - 8 - 6 - 16 + 24 = 0

So, the point (1,1,2) also lies on the sphere.

Step 2: Check if the gradients of the two surfaces at the point (1,1,2) are proportional.

The gradient of a surface F(x,y,z) = 0 at a point (x0,y0,z0) is given by the vector (F_x(x0,y0,z0), F_y(x0,y0,z0), F_z(x0,y0,z0)), where F_x, F_y, F_z are the partial derivatives of F with respect to x, y, z.

For the ellipsoid 3x^2 + 2y^2 + z^2 - 9 = 0, the gradient at (1,1,2) is (6x, 4y, 2z) = (6, 4, 4).

For the sphere x^2 + y^2 + z^2 - 8x - 6y - 8z + 24 = 0, the gradient at (1,1,2) is (2x - 8, 2y - 6, 2z - 8) = (-6, -4, -4).

The gradients are proportional (they are negatives of each other), so the ellipsoid and the sphere are tangent at the point (1,1,2).