The tangents at the points A(1, 3) and B(1, - | vedclass

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(D) The equation of the parabola is $y^{2} - 2y - 2x = 1$,which can be rewritten as $(y - 1)^{2} = 2(x + 1)$.Let the point of intersection of tangents

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$8$

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(D) The equation of the parabola is $y^{2} - 2y - 2x = 1$,which can be rewritten as $(y - 1)^{2} = 2(x + 1)$.Let the point of intersection of tangents at $A(x_{1}, y_{1})$ and $B(x_{2}, y_{2})$ be $P(h, k)$.The equation of the tangent at any point $(x_{1}, y_{1})$ on the parabola $y^{2} - 2y - 2x - 1 = 0$ is given by $yy_{1} - (x + x_{1}) - (y + y_{1}) - 1 = 0$.For point $A(1, 3)$,the tangent is $3y - (x + 1) - (y + 3) - 1 = 0$,which simplifies to $2y - x - 5 = 0$.For point $B(1, -1)$,the tangent is $-y - (x + 1) - (y - 1) - 1 = 0$,which simplifies to $-2y - x - 1 = 0$.Solving these two equations: $2y - x = 5$ and $-2y - x = 1$.Adding the equations gives $-2x = 6$,so $x = -3$.Substituting $x = -3$ into $2y - x = 5$ gives $2y + 3 = 5$,so $y = 1$.Thus,the point $P$ is $(-3, 1)$.The triangle $PAB$ has vertices $P(-3, 1)$,$A(1, 3)$,and $B(1, -1)$.The base $AB$ is a vertical line segment with length $|3 - (-1)| = 4$.The height of the triangle from $P$ to the line $AB$ (which is $x = 1$) is $|1 - (-3)| = 4$.Area of $	riangle PAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 4 	imes 4 = 8$ square units.

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

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