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A point moves such that the sum of the squares of its distances from 2 intersecting lines is constant(given that the lines are neither perpendicular nor do they make complementary angles). Prove that the locus is an ellipse.

A point moves such that the sum of the squares of its distances from 2 intersecting lines is constant(given that the lines are neither perpendicular nor do they make complementary angles). Prove that the locus is an ellipse.

Grade:upto college level

1 Answers

Ajay Verma
askIITians Faculty 33 Points
7 years ago
soln:

if we take 2 intersecting lines as :
L1: y = 0 ( for less calculation u can solve it by taking y= mx+c)
L2 : y= mx+c

and the point is P (h,k)
distance of point P(h,k) from L1= k ........(1)
distance of point P(h,k) from L2 = (y-m1x- c)/ (1+ m^2)^(1/2)

so k2 + [(k-mh- c)/ (1+ m2)1/2 ]2 = Q(constant)
(1+ m2)k2 + k2+ m2h2 + c2 - 2kmh - 2kc + 2mhc= Q( 1+ m2)
m2h2 + (2+ m2)k2- 2mkh + 2mch -2ck +c2= Q ( 1+ m2)
so locus would be

m2x2+ (2+ m2)y2- 2mxy + 2mcx -2cy +c2= Q ( 1+ m2)
its an eqn of conic..
Delta is not equal to zero... ( so its an non-degenerated conic)
and H < A.B
m2 < m2 ( 2+ m2)
m2 < 2m2 + m4 (its true...)
so the conic is an ellipse..

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