Consider a curve passing through (1,1) such that perpendicular distance of normal drawn at any point P on the curve from origin is equal to ordinate of the point P . If area enclosed by the curve is A , then find [2 A] . [Note: [k] denotes greatest integer less than or equal to k .]

Step-by-step solution

Equation of normal at P(x, y) aligned Y-y & = -1 m (X-x) m Y-m y & =-X+x x+m y-(x+m y) & =0 | m y+x 1+m^2 | & =y aligned m^2 y^2+x^2+2 x m y=m^2 y^2+y^2 m= y^2-x^2 2 x y gathered 2 x y d y d x =y^2-x^2 y^2=t 2 y d y d x = d t d x x d t d x =t-x^2 d t d x - t x =-x which is linear differential equation gathered aligned & I.F. =e^ - d x x = 1 x & t ( 1 x )=- 1 d x+C y^2 x =-x+C & aligned (1,1) C=-2y^2=-x^2+2 x x^2+y^2-2 x=0 (x-1)^2+y^2=1 Area bounded by the curve, A= [2 A]=[2 ]=6