If the tangent at (1, 1) on  $y^2=x{(2-x)}^2$ meets the curve again at P, then  P is

$2y\frac{dy}{dx}={(2-x)}^2-2x(2-x),so\left.\frac{dy}{dx}\right|{}_{{ }_{{ }_{{ }_{{ }_{(1,1)}}}}}=-\frac{1}{2}$

Therefore, the equation of tangent at (1, 1) is 

$\begin{array}{r}y-1=-\frac{1}{2}(x-1)\\ \Rightarrow y=\frac{-x+3}{2}\end{array}$

The intersection of the tangent and the curve is given by

$\begin{array}{l}(1/4)(-x+3{)}^2=x\left(4+x^2-4x\right)\\ \Rightarrow x^2-6x+9=16x+4x^3-16x^2\\ \Rightarrow 4x^3-17x^2+22x-9=0\\ \Rightarrow (x-1)\left(4x^2-13x+9\right)=0\\ \Rightarrow (x-1{)}^2(4x-9)=0\end{array}$

$\text{ Since }\mathrm{x}=1\text{ is already the point of }\underset{\_}{\text{ tangency, }\mathrm{x}}=9/4\text{ and }$

$y^2=\frac{9}{4}{\left(2-\frac{9}{4}\right)}^2=\frac{9}{64}\text{. }\mathrm{Thu}s\text{ the required point is (9/4, 3/8)}$