Question

Show that the ellipsoid 3x^2+2y^2+z^2= 9 and the sphere x^2+y^2+z^2-8x-6y-8z+24 = 0 are tangent to each other at the point (1,1,2), i.e., they have the same tangent plane there.

Answer 1

The equation of tangent plane is $z-z_0=h_{x}^{'}(x_0, y_0, z_0)(x-x_0)+h_{y}^{'}(x_0, y_0, z_0)(y-y_0).$

We have $F(x, y, z)=3x^2+2y^2+z^2- 9$ and $G(x, y, z)=x^2+y^2+z^2-8x-6y-8z+24$.

We must show that $F_{x}^{'}(1, 1, 2)=G_{x}^{'}(1, 1, 2)$ and $F_{y}^{'}(1, 1, 2)=G_{y}^{'}(1, 1, 2).$

$F_{x}^{'}=-\frac{\frac{\partial F(x, y, z)}{\partial x}}{\frac{\partial F(x, y, z)}{\partial z}}=-\frac{6x}{2z}=-\frac{3x}{z},$ $F_{x}^{'}(1, 1, 2)=-\frac{3}{2}.$

$F_{y}^{'}=-\frac{\frac{\partial F(x, y, z)}{\partial y}}{\frac{\partial F(x, y, z)}{\partial z}}=-\frac{4y}{2z}=-\frac{2y}{z},$ $F_{y}^{'}(1, 1, 2)=-1.$

$G_{x}^{'}=-\frac{\frac{\partial G(x, y, z)}{\partial x}}{\frac{\partial G(x, y, z)}{\partial z}}=-\frac{2x-8}{2z-8}=-\frac{x-4}{z-4},$ $G_{x}^{'}(1, 1, 2)=-\frac{3}{2}.$

$G_{y}^{'}=-\frac{\frac{\partial G(x, y, z)}{\partial y}}{\frac{\partial G(x, y, z)}{\partial z}}=-\frac{2y-6}{2z-8}=-\frac{y-3}{z-4},$ $G_{y}^{'}(1, 1, 2)=-1.$