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Analytical Geometry

Scorching goIITian

Joined:	7 Oct 2009
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14 Oct 2009 20:15:58 IST

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The ends A,B of a strait line line segment of a constant length 'c' slide upon the fixed rectangular

Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Analytical Geometry

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.

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Millind Gupta

Scorching goIITian

Joined: 7 Oct 2009

Posts: 227

14 Oct 2009 20:33:42 IST

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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

h^2/3+ k^2/3 = c^2/3

put h = x and k = y

x^2/3 + y^2/3 = c^2/3

^{Hence proved}

jee-do or die

Blazing goIITian

Joined: 15 Aug 2008

Posts: 584

31 Oct 2009 14:25:37 IST

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The ends A,B of a strait line line segment of a constant length 'c' slide upon the fixed r

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