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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