The question is a little obscure, so my solution is based on interpreting the question by insertion of certain assumed details.
The general equation of a parabola is a quadratic y=ax2+bx+c, where a>0 (upright parabola). This must first be written in vertex form:
y-k=a(x-h)2; y=ax2-2ahx+ah2+k; so b=-2ah and c=ah2+k, where (h,k) is the vertex.
Because the parabola has to have its vertex in the first quadrant h,k≥0. Since a>0 and h≥0, b≤0.
The parabola must not intersect the x-axis, so its zeroes must be complex, implying b2<4ac, that is:
4a2h2<4a(ah2+k), 4ah2<4ah2+4ak, 4ak>0⇒c>ah2 or c>b2/4a, and k>0.
The parabola must intersect both lines. To find x1, x2, x3 and x4 we have to solve two equations by substituting for y:
x=ax2+bx+c, ax2+(b-1)x+c=0 and 2x=ax2+bx+c, ax2+(b-2)x+c=0. We've already established b<0 and c>b2/4a.
So to intersect both lines, the parabola must reach far enough to intersect the line with the shallower gradient (y=x). This implies that 4ac≤(b-1)2, c≤(b-1)2/4a, therefore b2/4a<c≤(b-1)2/4a. Note that (b-1)2/4a<(b-2)2/4a because b<0.
The parabola intersects y=2x before it intersects y=x. ax2+(b-2)x+c=0 and ax2+(b-1)x+c=0 each have two solutions:
x1=(2-b-√((b-2)2-4ac))/2a, x3=(2-b+√((b-2)2-4ac))/2a;
x2=(1-b-√((b-1)2-4ac))/2a, x4=(1-b+√((b-1)2-4ac))/2a.
Note that 2-b and 1-b are positive numbers and 2-b>1-b because b is negative.
If D=|(x1-x2)-(x3-x4)|, that is D=|(x1-x3)-(x2-x4)|, then x1-x2=(1-√((b-2)2-4ac)+√((b-1)2-4ac))/2a;
x3-x4=(1+√((b-2)2-4ac)-√((b-1)2-4ac))/2a. D has not be defined, so I've assumed here that it's the difference between the distances between x1 and x2 and x3 and x4, where these values of x are the x-coordinates of the intersections from left to right. If this is not the correct assumption there's enough information in the calculations to make the relevant substitutions.
D=|(-√((b-2)2-4ac)+√((b-1)2-4ac)-√((b-2)2-4ac)+√((b-1)2-4ac))|/2a,
D=|√((b-1)2-4ac)-√((b-2)2-4ac)|/a=(√((b-2)2-4ac)-√((b-1)2-4ac))/a.
The conditions to be met are a>0 (given), b<0 and b2/4a<c≤(b-1)2/4a.
EXAMPLES: a=1, b=-1, then 1/4<c≤1; a=1, b=-2, then 1<c≤9/4. As b decreases (increases negatively) so c has a greater feasible range. As a increases (a>1) c has a smaller range. When 0<a<1 c has a greater range. If a=¼, b=-1, then 1<c≤4.
Let a=1, b=-1, c=1, then D=√5; let a=1, b=-2, c=2, then D=√8-1.