The ends A,B of a strait line line segment of a 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 X^2/3+y^2/3=c^2/3.
Actually this question was put to me on my nudgebook
thats why i am answering here bcoz here it is presentable and others can also be benefitted.
eqn of AB is x/a + y/b = 1 ----> bx + ay = ab
let coordinates of foot of perpendicular be (h,k)
sinc (h,k) lies on the line it should satisfy it bh + ak = ab --------(1)
(slope of AB) * (slope of perpendicular) = -1
so this gives a/b = (k - b)/(h - a) = ah - bk = a2 - b2 -----------(2)
& its given that line is of constant length c i.e. a2 + b2 = c2 -----------(3)
solve (1) & (2) to get the value of h & k
[ This can be done by multiplying (1) by a & (2) by b and add to get the value of h .similarly k can be found out]
h = a3 / (a2 + b2) & k = b3 / (a2 + b2)
using (3) we get h = a3 /c2 and k = b3/c2
a = (hc2)1/3 b = (kc2)1/3
put in (3)
(hc2)2/3 + (kc2)2/3 = c2
h2/3 + k2/3 = c2/3
put h = x and k = y
x2/3 + y2/3 = c2/3
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