The tangents at the points A(1, 3) and B(1, - 1) on the parabola y...

Given, ${y}^{2} - 2x - 2y = 1$

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

Tangents are drawn at \left(1, 3\right) \text{ and } \left(1, - 1\right)$, on plotting the diagram we get,</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/132fcf6d-2971-4341-bfe5-64ea0f7c71ce-image.png" style="width: 311px; height: 200px;" /></p><p>Now, tangent as$A: 2y - x - 5 = 0${using T = 0}$

Intersection with $y = 1$ is $x = - 3$

Hence, point $P$ is \left(- 3, 1\right)$</p><p>We can see diagram has symmetry</p><p>So, area of$∆ PAB = 2 × \dfrac{1}{2} × \left(1 - \left(- 3\right)\right) × \left(3 - 1\right)$</p><p>$= 8sq . units