If tangent lines are drawn from the point (-1 | vedclass

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

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(D) Given the parabola $y^2 = 4x$, we have $a = 1$. The point is $P(-1, 2)$.The equation of the chord of contact for $P(x_1, y_1)$ is $yy_1 = 2a(x + x_1)$.Substituting $x_1 = -1, y_1 = 2, a = 1$, we get $2y = 2(x - 1)$, which simplifies to $y = x - 1$ or $x - y - 1 = 0$.The length of the chord of contact $L$ is given by $\frac{\sqrt{(y_1^2 - 4ax_1)^3}}{2a} = \frac{\sqrt{(2^2 - 4(1)(-1))^3}}{2(1)} = \frac{\sqrt{(4 + 4)^3}}{2} = \frac{\sqrt{8^3}}{2} = \frac{16\sqrt{2}}{2} = 8\sqrt{2}$.The perpendicular distance $h$ from the point $(-1, 2)$ to the chord $x - y - 1 = 0$ is $h = \frac{|(-1) - (2) - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|-4|}{\sqrt{2}} = 2\sqrt{2}$.The area of the triangle formed by the tangents and the chord of contact is given by $\frac{h^3}{2a} = \frac{(2\sqrt{2})^3}{2(1)} = \frac{8 	imes 2\sqrt{2}}{2} = 8\sqrt{2}$.

If tangent lines are drawn from the point $(-1, 2)$ to the parabola $y^2 = 4x$, then the area of the triangle (in sq. units) formed by the chord of contact and the tangents drawn is: (in $\sqrt{2}$)

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