The tangents at the points $A\left(1,3\right)$ and $B\left(1,-1\right)$ on the parabola $y^2-2x-2y=1$ meet at the point $P$. Then the area (in ${\mathrm{unit}}^2$) of the triangle $PAB$ is:

Given,  $y^2-2x-2y=1$ 

$\Rightarrow {\left(y-1\right)}^2=2\left(x+1\right)$

Tangents are drawn at $\left(1,3\right) \& \left(1,-1\right)$, on plotting the diagram we get,

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Now, tangent as $A: 2y-x-5=0$ {using $T=0$}

Intersection with $y=1$ is $x=-3$

Hence, point $P$ is $\left(-3,1\right)$

We can see diagram has symmetry

So, area of $∆PAB=2\times \frac{1}{2}\times \left(1-\left(-3\right)\right)\times \left(3-1\right)$

$=8 \mathrm{sq}.\mathrm{units}$