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A variable line makes intercepts on the coordinate axes the sum of whose squares is constant and is equal to a^2. Find the locus of the foot of perpendicular from the origin to this line.
Dear Karthik,
eq of line is x/p + y/q = 1 where p , q are intercept on x. y axis
alos given, p2 + q2 = a2 eq 1
let foot of perpendicular is x1, y1
this point should lie on x/p + y/q = 1
so, x1/p + y1 /q =1
also (y1/x1 ) * ( - q/p ) = -1 eq2
find value of p, q from eq1 and eq2 and sunstitute in x1/p + y1 /q =1.
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