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(B) The equation of the pair of tangents from a point $(x_1, y_1)$ to the parabola $S: y^2 - 4x = 0$ is given by $SS_1 = T^2$.Here,$S = y^2 - 4x$,$S_1

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$6 \sqrt{2}$

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(B) The equation of the pair of tangents from a point $(x_1, y_1)$ to the parabola $S: y^2 - 4x = 0$ is given by $SS_1 = T^2$.Here,$S = y^2 - 4x$,$S_1 = (2)^2 - 4(-1) = 4 + 4 = 8$,and $T = y(2) - 2(x - 1) = 2y - 2x + 2$.Substituting these into the equation:$(y^2 - 4x)(8) = (2y - 2x + 2)^2$$8(y^2 - 4x) = 4(y - x + 1)^2$$2(y^2 - 4x) = (y - x + 1)^2$To find the intercept on the line $x = 2$,substitute $x = 2$ into the equation:$2(y^2 - 8) = (y - 2 + 1)^2$$2y^2 - 16 = (y - 1)^2$$2y^2 - 16 = y^2 - 2y + 1$$y^2 + 2y - 17 = 0$Let the roots be $y_1$ and $y_2$. Then $y_1 + y_2 = -2$ and $y_1 y_2 = -17$.The length of the intercept is $|y_1 - y_2| = \sqrt{(y_1 + y_2)^2 - 4y_1 y_2}$.$|y_1 - y_2| = \sqrt{(-2)^2 - 4(-17)} = \sqrt{4 + 68} = \sqrt{72} = 6 \sqrt{2}$.

Tangents are drawn from the point $(-1, 2)$ to the parabola $y^2 = 4x$. The length of the intercept made by these tangents on the line $x = 2$ is:

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