Ladders in the first quadrant

Question

Assume we're looking at the first quadrant in the plane with the positive $x$ and $y$ axes as boundaries. Look at a point $ (a,b) $ in the interior and consider ladders passing through this point, that is, line segments with one end on the $x$-axis, the other end on the $y$-axis and passing through $ (a,b) $. The question is this: what is the minimum length of such a ladder?

The answer is $ (a^{2/3} + b^{2/3})^{3/2} $ as can be found out with some easy calculus. But I'm wondering if there is a physical interpretation for such an answer, in particular for the exponent $ 2/3 $ and its reciprocal. The answer is neat, but the ladder which minimizes this does not have symmetrical $x$ and $y$ coordinate endpoints: $ x = a + (ab^2)^{1/3} $ and $ y = b + (a^2b)^{1/3} $, which is another curious case to ask this question.

Answer 1

For a physical interpretation, the given problem is equivalent to a rigid bar with fixed lenght constrained to move keeping end points on each of the axes and also costrained to keep the contact with the point $P=(a,b)$.

In this way, indeed, we set a stationary condition for the length of the line segment passing through $P=(a,b)$ which corresponds to its minimum value.

In such situation, for the first constraint, the instantaneous center of rotation must be located at $C=(x,y)$ and, for the second constraint, the line $CP$ must be orthogonal to the rigid bar.

Indeed, according to the following sketch

we obtain geometrically the conditions

$$\frac{x-a}{y-b}=\frac{y}{x}=\frac{b}{x-a}$$

which lead to the given solution.

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Remarks

1. The physical analogy explains also why point $P=(a,b)$ satisfies the astroid equation as noticed by mathlove in the comments.
2. Given $x$ and $y$ by this simple geometrical construction we can find $P=(a,b)$ on the line segment such that the given lenght is the minimum among all others line segments passing through $P=(a,b)$.

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Animation for the physical analogy