Find two points and one angle so that cone intersects the $xy$-plane in the parabola $y=x^2$ Find $\vec{p}\in \mathbb{R}^3$, $\vec{u}\in S^2$ and $\phi \in \left(0, \frac{\pi}{2}\right)$ such that the cone $C(\vec{p}, \vec{u}, \phi)$ intersects the $xy$-plane in the parabola $y=x^2$.
What I'm trying to do is the following:
(1) Pick point $\vec{p}=(0,0,h)$, pick $\phi = \frac{\pi}{4}$ (to be adjusted later as needed).
(2) The definition of the cone is as follows:
$$C=C(\vec{p}, \vec{u}, \phi)=\left\{\vec{x}\in \mathbb{R}^3:|(\vec{x}-\vec{p})\cdot \vec{u}|=|\vec{x}-\vec{p}|\cos\phi\right\},$$
so we have
$$\left((x,y,-h)\cdot (u,v,w)\right)^2=\left(x^2+y^2+h^2\right)\cos^2\phi=\frac12\left(x^2+y^2+h^2\right),$$
so that $$\left(xu+yv-hw\right)^2 = \frac12\left(x^2+y^2+h^2\right)$$
Now I'm lost. How can I now check whether or not my cone intersects the $xy$-plane in the parabola $y=x^2$?
I can probably substitute $y=x^2$ into $\frac12\left(x^2+y^2+h^2\right)$ to get $$\frac12\left(x^2+y^2+h^2\right)=\frac12\left(x^2+x^4+h^2\right)$$
But then what next?
 A: I'll show here a more geometric approach.
Let $V$ be the cone vertex (your $\vec p$), which by symmetry lies on the $yz$-plane, and $A=(0,a,0)$ the point where the axis of the cone meets $y$-axis (see diagram below). If lines $VB$ and $VC$ are the intersections of the cone with the $yz$-plane, then it is well known that one of them ($VC$ in the diagram) must be parallel to the $y$-axis, if the intersection between cone and $xy$-plane has to be a parabola. That entails $OV=OA=a$.
Let the plane through $A$ and perpendicular to $VA$ intersect lines $VB$ and $VC$ at $B$ and $C$, and intersect the parabola at $D$ and $E$: we have of course $AB=AC=AD=AE$. From triangle $ABV$ we get $AB=2a\sin\phi$, while the equation of the parabola gives $AD=\sqrt{a}$. Equating these gives:
$$
\sin\phi={1\over2\sqrt{a}},
\quad\hbox{with}\quad
a\ge{1\over4}.
$$
All cones satisfying this relation will thus intersect the $xy$-plane along the given parabola, every choice of $a$ or $\phi$ giving a different cone. 

EDIT 1.
From the discussion above, it is straightforward to find the coordinates of $V$:
$$
V=(0,-a\cos2\phi,\pm a\sin2\phi)
$$
where the $\pm$ accounts for the cone reflected about the $xy$-plane, which is also a valid solution. Inserting here $a=1/(4\sin^2\phi)$, and taking into account that $\vec p=V-O=V$, we get:
$$
\vec p=\left(0,{1-\cot^2\phi\over4},\pm {\cot\phi\over2}\right).
$$
Vector $\vec u$, on the other hand, is nothing but the direction of $A-V$, thus:
$$
\vec u=(0,\cos\phi,\mp\sin\phi).
$$
EDIT 2.
If you want to solve the question using your vector setting, then you can observe that by symmetry vectors $\vec p$ and $\vec u$ lie on the $yz$-plane. We can thus set $\vec p=(0,p_y,p_z)$, $\vec u=(0,\cos\alpha,\sin\alpha)$.
To find $p_y$, $p_z$ and $\alpha$ as a function of $\phi$, you can then apply the definition of cone to a point $\vec x$ of the given parabola, that is: $\vec x=(x,x^2,0)$. The equation you get is then:
$$
(x,x^2-p_y,-p_z)\cdot(0,\cos\alpha,\sin\alpha)=
\cos\phi\sqrt{x^2+(x^2-p_y)^2+p_z^2}.
$$
After squaring both sides you end up with a polynomial equation in $x$, which must hold for any value of $x$. That means the coefficients of powers of $x$ with the same exponent must be the same on both sides: that will allow you to find expressions for $p_y$, $p_z$ and $\alpha$.
