$\begin{array}{l}

y^{2}=x(2-x)^{2} \Rightarrow y^{2}=x^{3}-4 x^{2}+4 x \ldots \text { (i) } \\

\Rightarrow 2 y \frac{\mathrm{dy}}{\mathrm{dx}}=3 x^{2}-8 x+4 \\

\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3 x^{2}-8 x+4}{2 y} \\

\Rightarrow\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{P}}=\frac{3-8+4}{2}=-\frac{1}{2}

\end{array}$

$y^{2}=x(2-x)^{2} \Rightarrow y^{2}=x^{3}-4 x^{2}+4 x \ldots$ (i) $\Rightarrow 2 y \frac{\mathrm{dy}}{\mathrm{dx}}=3 x^{2}-8 x+4$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{3 x^{2}-8 x+4}{2 y}$ $\Rightarrow\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{P}}=\frac{3-8+4}{2}=-\frac{1}{2}$ $\therefore$ Equation of tangent at $\mathrm{P}$ is: $y-1=-\frac{1}{2}(x-1)$ $\Rightarrow x+2 y-3=0$ $\Rightarrow x+2 y-3=0$

Using $y=\frac{3-x}{2}$ in (i), we get: $\left(\frac{3-x}{2}\right)^{2}$

$\begin{array}{l}

=x^{3}-4 x^{2}+4 x \\

\Rightarrow 4 x^{3}-17 x^{2}+22 x-9=0 \text {...(ii) }

\end{array}$

which has two roots 1,1 (Because of (ii) being tangent at $(1,1)$ ).

Sum of 3 roots $=\frac{17}{4}$

$\begin{array}{l}

\therefore 3 \mathrm{rd} \text { root }=\frac{17}{4}-2=\frac{9}{4} \\

\text { Then, } y=\frac{3-\frac{9}{4}}{2}=\frac{3}{8} \\

\therefore \mathrm{Q} \text { is }\left(\frac{9}{4}, \frac{3}{8}\right)

\end{array}$