If the tangent at (1,1) on  ${\mathrm{y}}^2=\mathrm{x}(2-\mathrm{x}{)}^2$ 

meets the curve again at P, then P is

 $2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=(2-\mathrm{x}{)}^2-2\mathrm{x}(2-\mathrm{x}),\text{ so }{\left.\frac{\mathrm{dy}}{\mathrm{dx}}\right|}_{(1,1)}=\frac{1}{2}[1-2]=-\frac{1}{2}$ Therefore, the equation of tangent at (1,1) is $\mathrm{y}-1=-\frac{1}{2}(\mathrm{x}-1)\Rightarrow 2\mathrm{y}-2=-\mathrm{x}+1\Rightarrow \mathrm{y}=\frac{-\mathrm{x}+3}{2}$ The intersection of the tangent and the curve is given by $\left(\frac{1}{4}\right)(-\mathrm{x}+3{)}^2=\mathrm{x}\left(4+{\mathrm{x}}^2-4\mathrm{x}\right)$ 
$\begin{array}{l}\Rightarrow {\mathrm{x}}^2-6\mathrm{x}+9=16\mathrm{x}+4{\mathrm{x}}^3-16{\mathrm{x}}^2\Rightarrow 4{\mathrm{x}}^3-17{\mathrm{x}}^2+22\mathrm{x}-9=0\\ \Rightarrow (\mathrm{x}-1)\left(4{\mathrm{x}}^2-13\mathrm{x}+9\right)=0\Rightarrow (\mathrm{x}-1{)}^2(4\mathrm{x}-9)=0\end{array}$
Since x=1 is already the point of tangency $\mathrm{x}=\frac{9}{4}\text{ and }{\mathrm{y}}^2=\frac{9}{4}{\left(2-\frac{9}{4}\right)}^2=\frac{9}{64}$
Thus the required point is $\left(\frac{9}{4},\frac{3}{8}\right)$