The surface of a rigid body is an ellipsoid with equation $$ A x^{2}+B y^{2}+C z^{2}=k^{2} $$ in a rest frame with origin \(O\) at the centre of mass and axes aligned with the principal axes at the centre of mass; \(A, B\), and \(C\) are the principal moments of inertia at \(O\) and \(k\) is a constant. Show that if the body rotates freely about \(O\), with \(O\) fixed relative to an inertial frame, then the two tangent planes to the surface at the intersection points between the surface and the instantaneous axis are fixed relative to the inertial frame. Deduce that the body moves as if it were rolling between these two planes.

To find the intersection points, we first need to find the equation of the instantaneous axis. Since the rigid body is rotating freely about the origin, the instantaneous axis will be a line that passes through the origin and has a direction that coincides with the axis of rotation. Without loss of generality, let us consider the body rotating about the \(z\)-axis. So the equation of the instantaneous axis can be written as: $$x_0 = \alpha x, \ y_0 = \alpha y$$ with \(\alpha\) being a constant. Now, we need to find the intersection points by plugging the axis equation into the ellipsoid surface equation: $$A (\alpha x)^{2}+B (\alpha y)^{2}+C z^{2}=k^{2}$$ Simplify the equation to get: $$\alpha^{2}(A x^{2}+B y^{2})+C z^{2}=k^{2}$$ Notice that the intersection points will satisfy both the equation of the ellipsoid and the instantaneous axis equation.

To find the tangent planes that touch the ellipsoid surface at the intersection points, we will need to compute the gradient of the ellipsoid equation which gives us the normal vectors to the tangent planes. The gradient of the ellipsoid equation is given by: $$\nabla F(x,y,z) = \langle 2Ax, 2By, 2Cz \rangle$$ Now, we need to find the tangent plane's equation at the intersection points \((x_0,y_0,z_0)\). Using the gradient and the point-slope form of a plane, we get the equation of the tangent plane as: $$2A(x-x_0)x_0+2B(y-y_0)y_0+2C(z-z_0)z_0=0$$ As shown in the first step, \((x_0, y_0, z_0) = (\alpha x, \alpha y, z)\). Plugging these values and rearranging the terms, we get: $$2A(x-\alpha x)\alpha x+2B(y-\alpha y)\alpha y+2C(z-z)z=0$$ Simplifying, we have: $$2A(1-\alpha^2)x^2+2B(1-\alpha^2)y^2=0$$

Since we have the equation of the tangent planes, we can see that the coefficients do not depend on the coordinates \(x\), \(y\), or \(z\). This indicates that the tangent planes are fixed relative to the inertial frame.

The fact that the two tangent planes are fixed relative to the inertial frame suggests that as the body rotates, these planes do not move. Therefore, the points of contact between the surface and the instantaneous axis do not change in the inertial frame. This implies that the body moves as if it were rolling between these two planes.