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(D) The equation of the parabola is $y^2 = 4x$,so $a = 1$. The point is $(x_1, y_1) = (-1, 2)$. The equation of the chord of contact is $yy_1 = 2a(x +

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(D) The equation of the parabola is $y^2 = 4x$,so $a = 1$. The point is $(x_1, y_1) = (-1, 2)$. The equation of the chord of contact is $yy_1 = 2a(x + x_1)$,which gives $2y = 2(1)(x - 1)$,or $y = x - 1$. This can be written as $x - y - 1 = 0$. The length of the chord of contact is given by $L = \frac{2a}{m^2} \sqrt{1+m^2} \cdot |y_1^2 - 4ax_1|$,where $m$ is the slope of the chord. Alternatively,the area of the triangle formed by the tangents and the chord of contact is given by $\frac{(y_1^2 - 4ax_1)^{3/2}}{2a}$. Here,$y_1^2 - 4ax_1 = (2)^2 - 4(1)(-1) = 4 + 4 = 8$. Area $= \frac{(8)^{3/2}}{2(1)} = \frac{8 \sqrt{8}}{2} = 4 	imes 2 \sqrt{2} = 8 \sqrt{2}$.

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

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