3-D Geometry Problem. Find a curve which touches the straight line. If two perpendicular tangent planes to paraboloid  $x^{2}+y^{2}=2z$  internsects in a straight line in the plane $x=0$, obtain the curve to which the straight line touches.
I don't know how to proceed in this question as I am very poor in the 3-D Geomentry. Can Someone help me in this questions. I want a general approach to solve these type of questions.
 A: Let the tangent planes touch the surface at $(x', y',z')$ and $(x'',y'',z'')$ be
$$\left \{
  \begin{array}{rcl}
    x'x+y'y &=& z+z' \\
    x''x+y''y &=& z+z''
  \end{array}
\right.$$
respectively.
The line of intersection $(\xi, \eta, \zeta)$ is
$$(\xi, \eta, \zeta)=
\left(
  \frac{\begin{vmatrix} \zeta+z' & y' \\ \zeta+z'' & y'' \end{vmatrix}}
       {\begin{vmatrix} x' & y' \\ x'' & y'' \end{vmatrix}},
  \frac{\begin{vmatrix} x' & \zeta+z'  \\ x'' & \zeta+z'' \end{vmatrix}}
       {\begin{vmatrix} x' & y' \\ x'' & y'' \end{vmatrix}},
  \zeta
\right)$$
For $\xi=0$, $$ \left \{ \begin{align*} y'&=y''\\ z'&=z'' \end{align*} \right.$$
Note the normals for the tangent planes are $(x',y',-1)$ and $(x'',y'',-1)$.
For perpendicular tangent planes,
\begin{align*}
  x'x''+y'y''+1 &= 0 \\
  x'x'' &= -(1+y'^2)
\end{align*}
Also, $x'x''$ is the product of roots of $x^2+y'^2 = 2z'$,
$$x'x'' = -(1+y'^2) = y'^2-2z'$$
Let $t=y'$, then $(x',y',z')$ and $(x'',y'',z'')$ are given by
$$\begin{pmatrix} \pm \sqrt{1+t^2} \\ t \\ t^2+\frac{1}{2} \end{pmatrix}$$
Now
$$(\xi, \eta, \zeta)=
\left(
  0, \frac{t^2+\frac{1}{2}+\zeta}{t}, \zeta
\right)$$
and the family of the intersection,
$$F(Y,Z;t)=tY-Z-t^2-\frac{1}{2}=0$$
$$\frac{\partial F}{\partial t}=0 \implies t=\frac{Y}{2}$$
The envelope is $$
\left \{
  \begin{align*}
    Y^2 &= 4Z+2 \\
    X &= 0
  \end{align*}
\right.$$
A: Consider the two points at which the tangent planes touch the paraboloid:
$\left(x_1, y_1,\frac{x_1^2+y_1^2}{2}\right)$ and $\left(x_2, y_2,\frac{x_2^2+y_2^2}{2}\right)$
The tangent planes to these points are:
$z-\frac{x_1^2+y_1^2}{2}=x_1(x-x_1)+y_1(y-y_1)$ and $z-\frac{x_2^2+y_2^2}{2}=x_2(x-x_2)+y_2(y-y_2)$. 
(This is from the equation $z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$)
The intersection of these two lines must be parallel to the vector $( x_1,y_1,1)\times( x_2,y_2,1)$ whose $x$-component equal to $y_1-y_1$. Since the $x$-component must be equal to $0$, we conclude that $y_1-y_2=0$. In other words, $y_1=y_2$. In order to avoid confusion, let us define a new variable $y_3=y_1=y_2$.
Now, let's find a point of intersection by setting these two equations equal to each other:
$z=x_1(x-x_1)+y_1(y-y_1)+\frac{x_1^2+y_1^2}{2}$ and $z=x_2(x-x_2)+y_2(y-y_2)+\frac{x_2^2+y_2^2}{2}$
$$x_1(x-x_1)+y_3(y-y_3)+\frac{x_1^2+y_3^2}{2}=x_2(x-x_2)^2+y_3(y-y_3)+\frac{x_2^2+y_3^2}{2}$$
$$x_1\cdot x-\frac{x_1^2}{2}=x_2\cdot x -\frac{x_2^2}{2}$$
$$(x_1-x_2)x=\frac{x_1^2-x_2^2}{2}$$
$$x=\frac{x_1+x_2}{2}$$
Since the $x$ coordinate of any point of intersection must be $0$, we conclude that $x_1=-x_2$. Again, in order to avoid confusion, we create a new variable: $x_3=x_1=-x_2$.
Since these planes are perpendicular, $(x_1,y_1,1)\cdot(x_2,y_2,1)=0$. In other words, $x_3^2-y_3^2=1$ 
Now we can finally find the equation of the intersection line. We already know the the $x$-coordinate will be $0$, so we set $x$ equal to $0$ (we can do this to any of the two equations, the result will be the same):
$$z=-x_3^2+y_3(y-y_3)-\frac{x_3^2+y_3^2}{2}$$
$$z=y_3\cdot y - y_3^2-\frac{1}{2}$$
By rearranging the terms, we get a quadratic in $y_3$:
$$y_3^2-y\cdot y_3+z+\frac{1}{2}$$
Two find the envelope of this family of curves, all we have to do is set the discriminant to $0$: 
$$y^2-4z-2=0$$
$$z=\frac{y^2-2}{4}$$
The final curve is 
$$\begin{cases}z=\frac{y^2-2}{4}\\x=0\end{cases}$$
