Parabola Investigation IB HL Portfolio (can't get part 3)

Question

1. Consider the parabola
2 2 y (x 3) 2 x 6x 11 and the
lines y x and y 2x .
Using technology find the four
intersections illustrated on the right.
Label the x-values of these intersections
as they appear from left to right on the
x-axis as 1 2 3 4 x , x , x , and x .
Find the values of 2 1 x x and 4 3 x x
and name them respectively and L R S S.
Finally, calculate L R D S S .
2. Find values of D for other parabolas of the form 2 y ax bx c , a 0, with vertices in quadrant 1,
intersected by the lines y x and y 2x . Consider various values of a, beginning with a=1.
Make a conjecture about the value of D for these parabolas.
3. Investigate your conjecture for any real value of a and any placement of the vertex. Refine your
conjecture as necessary, and prove it. Maintain the labeling convention used in parts 1 and 2 by
having the intersections of the first line to be 2 x and 3 x and the intersections with the second line to
be 1 x and 4 x .

Answer 1

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.