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

2ydydx=(2−x)2−2x(2−x),so dydx|(1,1)=−12. Therefore , the equation of tangent at (1,1) is y−1=−12(x−1)                                                       ⇒            y=−x+32  The intersection of the tangent and the curve is given by              (1/4)(−x+3)2=x(4+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=94(2−94)2=964. Thus the required point is (9/4,3/8).