If tangents are drawn from the point (-1, 2) | vedclass

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(D) For the 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

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$8$

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(D) For the 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 $y(2) = 2(1)(x - 1)$, which simplifies to $y = x - 1$ or $x - y - 1 = 0$.The points of contact are found by solving $y^2 = 4x$ and $x = y + 1$, leading to $y^2 - 4y - 4 = 0$.The length of the chord of contact $L$ is given by $\frac{\sqrt{(y_1^2 - 4ax_1)^3}}{a} = \frac{\sqrt{(2^2 - 4(1)(-1))^3}}{1} = \sqrt{8^3} = 16\sqrt{2}$.The distance $d$ from the point $(-1, 2)$ to the line $x - y - 1 = 0$ is $d = \frac{|-1 - 2 - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.The area of the triangle formed by the tangents and the chord of contact is given by $\frac{1}{2} 	imes L 	imes d = \frac{1}{2} 	imes 16\sqrt{2} 	imes 2\sqrt{2} = 8 	imes 2 	imes 2 = 32$ (Wait, re-evaluating: The formula for the area of the triangle formed by tangents from $(x_1, y_1)$ to $y^2=4ax$ is $\frac{(y_1^2 - 4ax_1)^{3/2}}{2a} = \frac{(4+4)^{3/2}}{2} = \frac{8\sqrt{8}}{2} = 4(2\sqrt{2}) = 8\sqrt{2}$).

If tangents are drawn from the point $(-1, 2)$ to the parabola $y^2 = 4x$, what is the area of the triangle formed by the chord of contact and the tangents (in $\sqrt{2}$)?

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