Finding the smallest square inside a parabola.

Question

I just thought of a problem earlier today, but wanted to know if there was an easier way of acquiring the answer.

Say I have a standard parabola $y=x^2$ with 3 points on it $P,Q,R$ and another point $S$ such that $PQRS$ is a square.

Now I can clearly see that if I have the origin $(1,1), (-1,1)$ and $(0,2)$, then I form a square with area 2. However, this may not necessarily be the smallest square.

My idea for finding the square with the smallest area is to parameterise the parabola with 3 points $(t,t^2)$ where $t=p,q,r$, then to find the area of that triangle (perhaps using the determinant) and then minimise that quantity with the constraint that the interval $PQ=PR$ and that $PQ \perp PR$.

However, this seems too tedious. Is there a better way perhaps?

Answer 1

You could minimize the Euclidean distance between two intersection points of an arbitrary secant line.

Subject to constraints that guarantee the following:

1. The distance between these two points is the same as the distance between the two intersections, with the parabola, of a line normal to the secant.
2. The two lines and parabola must all intersect at one point and no more than one point.
3. The two lines are perpendicular.
4. All four points are distinct.

Perhaps this uses more fundamental concepts than your solution.