Question:
Consider (double) infinite cone for which boundary is defined by the following implicit equation for $\vec{x} = (x,y,z) \in \mathbb{R}^3$.
$$ F(\vec{x}) = F(x,y,z) = x^2 + y^2 - z^2 \tan^2 \theta = 0 $$
So, boundary is a set of all points $\vec{x}$ that satisfy $F(\vec{x}) = 0 $. Then, define inside of the cone as set of all points $\vec{x}$ with $F(\vec{x}) < 0 $.
Now, define line to be set of all points $\vec{x}$ such that given fixed $\vec{x}_0 = (x_0, y_0, z_0)$ and $\vec{v} = (v_x, v_y, v_z)$, it can be written as $\vec{x} = \vec{x}_0 + \vec{v}t $, where $t \in \mathbb{R}$.
If there is a point on the line that is inside the cone (I believe without any loss of generality one can take it to be $\vec{x}_0$, if not - reparametrize), is there a point in the line that is in the boundary, in other words, $\vec{x}$ that is both in the line and that has $F(\vec{x})=0$? Geometrically, my intuition tells me that such a result is true and could be proven.
My approach:
Using appropriate rotation around $z$ axis of both $\vec{x}_0$ and $\vec{v}$ one can obtain equivalent problem for which $\vec{v} = (v_x, 0, v_z)$. In that case, if $\vec{x}$ is on the line and has $F(\vec{x}) = 0$ then it amounts to $\vec{F}(\vec{x}_0 + \vec{v}t) = 0$ for at least one value of $t$. Using implicit function $F$, one can write this as a second order polynomial in $t$.
$$ F(\vec{x}_0 + \vec{v}t) = t^2 (v_x^2 - v_z^2 \tan^2 \theta) + 2t (xv_x - zv_z \tan^2 \theta) + (x^2+y^2 - z^2\tan^2\theta) = 0$$
This is in the form $at^2 + 2bt + c = 0$ and it has solutions if and only if $\Delta = b^2 - ac \geq 0$. Using our results of computations, this is equivalent to the following.
$$ \Delta = (xv_x - zv_z \tan^2 \theta)^2- (v_x^2 - v_z^2 \tan^2 \theta)(x^2+y^2 - z^2\tan^2\theta) $$
Here, I note that in the case of $ v_x^2 \geq v_z^2 \tan^2 \theta $, the second term is positive as $\vec{x}_0$ is inside the cone, and $\Delta \geq 0$.
Now, consider a case where $ v_x^2 < v_z^2 \tan^2 \theta $. One simple case of this would be that $v_x = 0$. In that case, we have the following.
$$ \Delta = z^2 v_z^2 \tan^4 \theta + v_z^2 \tan^2 \theta(x^2+y^2 - z^2\tan^2\theta) = v_z^2 \tan^2 \theta (x^2 + y^2) \geq 0 $$
So, now I am just left with the case of $ v_x^2 < v_z^2 \tan^2 \theta $ and $v_x \neq 0$. Unfortunately, after trying usual inequalities as Cauchy-Schwarz, triangle inequality, and trying to expand (or factor) a square of sum, I could not get anything useful.
I really would appreciate your help and advice!
You want to show that for $v_x^2 \lt v_z^2 \tan^2\theta$, $\Delta \ge 0$. Expanding and cancelling terms, this is equivalent to $$ -2xzv_x v_z \tan^2\theta \ge y^2 v_x^2 -\tan^2\theta \ (x^2 v_z^2 +y^2 v_z^2 +z^2 v_x^2)\\ \iff \tan^2\theta \big[ (xv_z +zv_x)^2 +y^2 v_z^2 \big] \ge y^2 v_x^2 $$ But this is true as the LHS is $$\ge y^2 v_z^2 \tan^2\theta \gt y^2 v_x^2 $$ by the starting assumption.