Intersection of a line segment and a paraboloid in 3D

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Suppose I have a line segment $L$ in 3D: $$x=a_1(1-t)+b_1t$$ $$y=a_2(1-t)+b_2t$$ $$z=(a_1^2+a_2^2-k_1^2)(1-t)+(b_1^2+b_2^2-k_2^2)t$$

Because $L$ is line segment then $0\leq t\leq 1$.

And defining paraboloid $P$ in 3D: $$P:z=2x^2+2y^2-1$$

Where $a_1,a_2,b_1,b_2,k_1,k_2$ are all variables and $k_1$ and $k_2$ are positive numbers.

I want to put some constraints on these variables such that $L$ and $P$ intersect or do not intersect.

I know by substituting $L$ into $P$ and then solving for $t$ where $0\leq t\leq 1$ is the solution of the intersection. But still i could not figure out about the constraints i am looking for.

Thanks for any suggestions.

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If you only want to know if they intersect or not, you don't need to figure out where do they intersect...

The endpoints of your segment are at $(a_1, a_2, a_1^2+a_2^2-k_1^2)$ and $(b_1, b_2, b_1^2+b_2^2-k_2^2)$. If both endpoints are above, or both below the paraboloid, then the segment does not intersect the segment. For instance, the first endpoint is above the paraboloid if $a_1^2+a_2^2-k_1^2 > 2a_1^2+2a_2^2-1$. Working it out, both endpoints will be above the paraboloid if

$$k_1^2<1-a_1^2-a_2^2,$$ $$k_2^2<1-b_1^2-b_2^2,$$

You will have an intersection if only one of the above inequalities holds.