If the line x a + y b =1 moves in such a way that 1 a^2 + 1 b^2 = 1 c^2 where, c is a constant, then the locus of the foot of perpendicular from the origin on the straight line is

Step-by-step solution

Variable line is x a + y b =1 ....(i) Any line perpendicular to (i) and passing througn tne oriain will be x b - y a =0 ...(ii) Now foot of the perpendicular from the origin to line (i) is the point of intersection (i) and (ii) Let it be P( , ) , then a + b =1 ...(iii) and b - a =1 ...(iv) Squarring and adding Eqs. (iii) and (iv), we get ^2 ( 1 a^2 + 1 b^2 )+ ^2 ( 1 b^2 + 1 a^2 )=1 . ( ^2+ ^2 ) 1 c^2 =1 Hence, the locus of P( , ) is x^2+y^2=c^2 which is a circle.