Function for length of straight line in a parabola If you map the equation  $y=x^2$ you get a parabola.
If one imagines that the parabola is a pipe, and that pipe has a diameter $t$ 
in relation to $t$ and $x$ how far can a straight line (starting from the middle) go if it extends the furthest possible in one direction as shown below- what is the function $f$($t$,$x$).

my question is with a given value for $t$ and $x$ what is the length of the line
edit: the central line is $y=x^2$ (the center of the 'pipe' if you will)but the upper and lower lines are at distance  ±t/2 therefore not parabolas, see this diagram 

 A: Hint:
This figure (from here) illustrate how we can find the parametric equation of the curve  parallel to a given curve.
In your case the red curve is the central parabola of equation $Y=f(X)=X^2$, $(x,f(x))$ is a point ($A$ in the figure) on this curve and $f'(x)=2x$ in your case is the derivative at this point.
Finally, if you have the equation of the line  (in your post is not so clear how this line is defined) you can find the intersection with the two parallels curves and the distances between them.
A: Not having your answer to my comment above, let's suppose that the parabola $y=x^2$ be the axis and that the inner and outer lines be at distance $\pm t/2$ measured (with sign) along the normal to the parabola.
Take a point $P$ on the parabola and the corresponding points $P_i$ and $P_o$ where the normal in $P$ crosses the flanking curves.
Then in the respective points the three curves shall clearly have the same curvature center.
Thus the respective tangents shall be three parallel lines distantiated by $t/2$.
Mark the points  where the tangent to the inner curve crosses the outer one:
then these points will distant $t/2$ from the parabola and $t/2$ from the line tangent to the parabola in $P$.  
Can you take on from here ? 
P.S.: I saw now your edit. It seems that now the final conditions become :
 $\cdots$ at distance $0$ or $t/2$ from the parabola, and $t/2$ from the tangent. 
A: I am giving you the solution for a general parabola $y^2=4ax $

For a general parabola $y^2=4ax$, we have to find the length of a straight line along it.

First we will take their points of intersection.

Equation of line is $y=mx+c$

$(mx+c)^2 =  4ax$

$m^2 x^2 + (2mc-4a)x + c^2=0$

$D>0$

The line will intersect the parabola at 2 points $A$($x_1,y_1$) and $B$($x_2,y_2$)

$AB= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$

$= |x_2-x_1|\sqrt{1+m^2}$

