The end A, B of a straight line segment of constant length c slide upon the fixed rectangular axes OX, OY respectively. If the rectangle OAPB be completed, then show that the locus of the foot of the perpendicular drawn from P to AB is 
+ 
= 
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. Here equation of AB is 
+ 
= 1
⇒ x sin α + y cos α = c sin α cos α …(1)
Equation of PN (perpendicular to AB and through P)
y – c sin α = cot α (x – c cos α )
⇒ x cos α – y sin α = c(cos 2 α – sin 2 α ) …(2)
N is intersection point of (1) and (2)
Multiplying (1) by sin α and (2) by cos α and subtracting.
We get, x = c cos 3 α , y = c sin 3 α
∴ cos α = 
, sin α = 
∴ locus of (x, y) is 
+ 
= 1
⇒ x 2/3 + y 2/3 = c 2/3
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