The tangents at the point A(1, 3) and B(1, –1) on the parabola y² – 2x – 2y = 1 meet at the point P. Then the area (in unit²) of the triangle PAB is :-

$$\begin{gathered} \text{ Let }P\text{ be }\left(\alpha ,\beta \right) \\ \text{ Chord of contact }\mathrm{AB}\to {\mathrm{S}}_1=0 \\ \begin{array}{l}\Rightarrow \beta y-(x+\alpha )-(y+\beta )=1\\ \Rightarrow x+y(1-\beta )+\alpha +\beta +1=0\\ \text{ Equation of }AB\to x=1\\ comparing both the equations of AB\to \beta =1\text{ and }\alpha +\beta +1=-1\Rightarrow a=-3\\ \Rightarrow \mathrm{P}\left(-3,1\right),\mathrm{A}(1, 3) \mathrm{and} \mathrm{B}(1, –1) \\ \mathrm{\Delta PAB}=\frac{1}{2}\left|\begin{array}{ccc}1& -1& 1\\ 1& 3& 1\\ -3& 1& 1\end{array}\right|=\frac{1}{2}\left(2+4+10\right)=8\end{array} \end{gathered}$$