Line segment AB of fixed length c slides between coordinate axes such that its ends A and B lie on the axes. If O is origin and rectangle OAPB is completed, then show that the locus of the foot of the perpendicular drawn from P to AB is `x^((2)/(3)) + y^((2)/(3)) = c^((2)/(3)).`

Explanation

To solve the problem step by step, we will derive the equation of the locus of the foot of the perpendicular drawn from point P to line segment AB, which lies between the coordinate axes.

### Step 1: Understand the Geometry
Let points A and B lie on the y-axis and x-axis respectively. Since the length of segment AB is fixed at \( c \), we can denote the coordinates of A as \( (0, A) \) and B as \( (B, 0) \). The relationship between A and B due to the fixed length is given by:
\[ AB = \sqrt{A^2 + B^2} = c \]
Thus, we have:
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