Intersection of two perpendicular tangent planes to the paraboloid $x^2 + y^2 = 2z$ Two perpendicular tangent planes to the paraboloid 
$x^2 + y^2 = 2z $
intersect in a straight line in the plane $ x = 0$
Obtain the curve to which this straight line touches.
I tried this by forming equations of planes
$ l_1x + m_1y +n_1z = -(l_1^2 + m_1^2)/2n_1 $
$ l_2x + m_2y +n_2z = -(l_2^2 + m_2^2)/2n_2 $
But I couldn't proceed further.
 A: The paraboloid can be parametrized by $$(u,v) \mapsto r =\left(u, v, \frac{u^2+v^2}2\right)$$
So $$\frac{\partial r}{\partial u} = (1, 0, u)\\\frac{\partial r}{\partial v} = (0, 1, v)$$
are two tangent vectors in the tangent plane. Their cross product $(u,v,-1)$ is therefore orthogonal to the tangent plane. For the point $(x_0, y_0, z_0)$ on the paraboloid, $u = x_0, v = y_0$, so the plane normal is $(x_0, y_0, -1)$. Since the point must be on the plane, the plane equation is $$(x_0, y_0, -1)\cdot (x, y, z) = (x_0, y_0, -1)\cdot(x_0,y_0,z_0)$$
which expands to $$x_0x + y_0y - z = x_0^2 + y_0^2 - z_0$$
or $$x_0x + y_0y - z = z_0$$
since $x_0^2 + y_0^2 = 2z_0$.
This tangent plane intersects the plane $x = 0$ in the line $y_0y = z + z_0$. The tangent plane at a second point $(x_1, y_1, z_1)$ intersects in the line $y_1y = z+z_1$. These two equations can only be equations of the same line if $y_1 = y_0$ and $z_1 = z_0$. (From the equation for the paraboloid it follows that $x_1 = \pm x_0$, and since the points need to be distinct, we have $x_1 = -x_0$.)
So the points of tangency must be $(x_0, y_0, z_0)$ and $(-x_0, y_0, z_0)$, where $x_0^2 + y_0^2 = 2z_0^2$, and the line of intersection must be $$x = 0;\quad z = y_0y - z_0$$
And as I've already mentioned, this line "touches" infinitely many curves, so I have no idea what curve they are after.
