Q.214. Locus of a point.

Question 214.
Ends A and B of a straight line segment of constant length C slides on fixed axis OY and OX. If rectangle OAPB be completed , then show that the locus of the foot of perpendicular  from P on AB is x^(2/3) + y^(2/3) = C^(2/3).

Answer 214.
Let A = (0, a) and B = (b, 0)
=> P = (b, a)
and AB^2 = a^2 + b^2
=> C^2 = a^2 + b^2

The equation of line AB is
x/b + y/a = 1
=> ax + by = ab ... (1)

The equation of a line perpendicular to AB is of the form
bx - ay = k
If it passes through P (b, a)
b^2 - a^2 = k
=> the equation of the line through P perpendicular to AB is
bx - ay = b^2 - a^2 ... (2)

Solving equations (1) and (2), the coordinates of the foot of perpendicular from P on AB are
x = b^3 / (a^2 + b^2) and y = a^3 / (a^2 + b^2)
=> x^(2/3) + y^(2/3)
= b^2 / (a^2 + b^2)^(2/3) + a^2 / (a^2 + b^2)^(2/3)
= (a^2 + b^2) / (a^2 + b^2)^(2/3)
= (a^2 + b^2)^(1/3)
= (C^2)^(1/3)
= C^(2/3)

Thus, the locus of the foot of perpendicular drawn from P on AB is x^(2/3) + y^(2/3) = C^(2/3).

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