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

As shown in the figure, is on x-axis and B is on y-axis.
AB=c (given)
Given that OAPB is a rectangle.
Let `angle BAO = alpha`
`therefore OA = c "cos " alpha and OB= " c sin "alpha`
`therefore P-= (c "cos " alpha,c "sin " alpha)`
Equation of AB is
` (x)/(c " cos " alpha) + (y)/(c "sin " alpha) = 1`
` or x "sin " alpha + y "cos"alpha = c " sin"alpha "cos" alpha " "(1)`
Equation of PN (perpendicular from P to AB) is
`y-c sin alpha = cot a (x-c cos alpha)`
`rArr x cos alpha - y sin alpha = c(cos^(2) alpha -sin^(2) alpha) " "(2)`
N is point of intersection of (1) and (2).
Solving (1) and (2), we get
`x = c cos^(3) alpha, y = c sin^(3) alpha`
`therefore cos alpha = ((x)/(c))^(1//3), sin alpha = ((y)/(c))^(1//3)`
Squaring and adding, we get
`((x)/(c))^(2//3) + ((y)/(c))^(2//3) = 1`
` or x^(2//3) + y^(2//3) = c^(2//3), "which is the required locus"`