From the point (-1, 2),tangent lines are draw | vedclass

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(C) For a parabola $y^2 = 4ax$,where $a = 1$,the chord of contact from point $(x_1, y_1) = (-1, 2)$ is given by $yy_1 = 2a(x + x_1)$.Substituting the

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$8\sqrt{2}$

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(C) For a parabola $y^2 = 4ax$,where $a = 1$,the chord of contact from point $(x_1, y_1) = (-1, 2)$ is given by $yy_1 = 2a(x + x_1)$.Substituting the values,we get $2y = 2(x - 1)$,which simplifies to $y = x - 1$ or $x - y - 1 = 0$.To find the points of contact $P$ and $Q$,we substitute $x = y + 1$ into $y^2 = 4x$:$y^2 = 4(y + 1) \implies y^2 - 4y - 4 = 0$.The roots are $y = \frac{4 \pm \sqrt{16 + 16}}{2} = 2 \pm 2\sqrt{2}$.Thus,the points are $P(3 + 2\sqrt{2}, 2 + 2\sqrt{2})$ and $Q(3 - 2\sqrt{2}, 2 - 2\sqrt{2})$.The length of the chord of contact $PQ = \sqrt{(4\sqrt{2})^2 + (4\sqrt{2})^2} = \sqrt{32 + 32} = 8$.The perpendicular distance $p$ from the point $(-1, 2)$ to the line $x - y - 1 = 0$ is $p = \frac{|-1 - 2 - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.The area of the triangle formed by the tangents and the chord of contact is given by $\frac{1}{2} 	imes PQ 	imes p = \frac{1}{2} 	imes 8 	imes 2\sqrt{2} = 8\sqrt{2}$.

From the point $(-1, 2)$,tangent lines are drawn to the parabola $y^2 = 4x$. The area of the triangle formed by the chord of contact and the tangents is given by:

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