If the tangent at (1, 1) on  y2 =  x(2−x)2 meets the curve again at P, then  P is

2y  dydx   =   (2−x)2−2x(2−x),  so  dydx     (1, 1)  =  −12Therefore, the equation of tangent at (1, 1) is y−1=−12(x−1)⇒ y=−x+32The intersection of the tangent and the curve is given by(1/4)(−x+3)2=x4+x2−4x⇒x2−6x+9=16x+4x3−16x2⇒4x3−17x2+22x−9=0⇒(x−1)4x2−13x+9=0⇒(x−1)2(4x−9)=0 Since x=1 is already the point of  tangency, x_=9/4 and y2=942−942=964. Thus  the required point is (9/4,  3/8)