Hokkaido University 2003: Parabolas and Loci

Problem

Suppose that the parabolas $${A:\ y=x^2}$$ and $${B:\ y=-(x-a)^2+b}$$ on the $${xy}$$ plane intersect at two distinct points P$${(x_1,\ y_1)}$$ and Q$${(x_2,\ y_2)}$$ ($${x_1>x_2}$$). When $${a}$$ and $${b}$$ vary while satisfying $${x_1-x_2=2}$$, illustrate the region through which the line PQ passes.
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The actual problem includes an introductory part: '(1) When $${x_1-x_2=2}$$ holds, express $${b}$$ in terms of $${a}$$.' Without this introduction, it would likely be quite difficult, as it reduces the number of variables by one. For the equation of the line PQ, even without finding the coordinates of the intersection points, one could use the method that 'the equation of a curve passing through the intersection of two curves $${f(x,y)=0}$$ and $${g(x,y)=0}$$ is $${f(x,y)+kg(x,y)=0}$$.' By setting $${k=-1}$$, we get $${y=ax-a^2+b}$$, but one might get stuck on what to do next. If $${b}$$ is expressed in terms of $${a}$$, it can be treated as a quadratic equation in terms of $${a}$$, allowing the use of the real root condition. This is the standard method for finding an envelope.

Now, setting aside the calculations, let's 'appreciate' the problem. It creates a quite beautiful figure.

Opening the link will take you to the following screen.

Dragging the green point at the vertex of the parabola changes the values of $${a}$$ and $${b}$$. However, as mentioned above, the value of $${b}$$ is determined by the value of $${a}$$, so it cannot move just anywhere.
Let's drag the green point while focusing on the line PQ. Can you visualize the region that PQ passes through?

Since the value of $${b}$$ is determined by the value of $${a}$$, the green point does not follow the mouse pointer perfectly, and it may sometimes go off-screen.

If you release the mouse button here, you won't be able to track the movement anymore, so in that case, press the 'Reset' button. It will return to the initial state.
Once you have grasped the movement of the line PQ, click the 'Show PQ Trace' button. The path of its movement will remain displayed.

This will reveal the envelope. It looks like a parabola, but is it?
Click the 'Show Region' button.

Calculating it, the envelope becomes $${y=x^2+1}$$.
Click the 'Show PQ Trace' button to clear the locus of PQ. It is a quite beautiful, symmetrical figure.

Although it is not in the problem, do you notice anything while moving PQ with the region displayed? The envelope $${y=x^2+1}$$ and the line PQ are tangent, and it seems that the point of tangency is the midpoint of PQ. Let's calculate and verify this.

*The figures are created using Cinderella (CindyJS).