Thank you for reporting, we will resolve it shortly
Q.
The tangents at the point $A (1,3)$ and $B (1,-1)$ on the parabola $y^2-2 x-2 y=1$ meet at the point $P$. Then the area (in unit $^2$ ) of the triangle PAB is :-
Both point $A (1,3), B (1,-1)$ lies on the parabola
$y^2-2 y-2 x-1=0$
Fquation of tangent aty $A(1,3)$ is $T=0$
$x-2 y+5=0$
and equation of tangent at $B (1,-1)$ is $T =0$
$x +2 y +1=0$
So point $P$ is $(-3,1)$
$\Rightarrow A =\frac{1}{2} \begin{vmatrix}1 & 3 & 1 \\1 & -1 & 1 \\-3 & 1 & 1\end{vmatrix}=8$