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(C) The equation of the parabola is $y^2 = 4x$,so $a = 1$. The point $(-2, -1)$ lies outside the parabola because $(-1)^2 - 4(-2) = 1 + 8 = 9 > 0$. Th

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$\frac{-3}{4}$

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(C) The equation of the parabola is $y^2 = 4x$,so $a = 1$. The point $(-2, -1)$ lies outside the parabola because $(-1)^2 - 4(-2) = 1 + 8 = 9 > 0$. The equation of a tangent to $y^2 = 4ax$ with slope $m$ is $y = mx + \frac{a}{m}$. Substituting $a = 1$ and the point $(-2, -1)$,we get $-1 = -2m + \frac{1}{m}$. Multiplying by $m$,we get $-m = -2m^2 + 1$,which simplifies to $2m^2 - m - 1 = 0$. Factoring the quadratic,$(2m + 1)(m - 1) = 0$,so the slopes are $m_1 = 1$ and $m_2 = -1/2$. The angle $	heta$ between the tangents is given by $	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}|$. $	an 	heta = |\frac{1 - (-1/2)}{1 + (1)(-1/2)}| = |\frac{3/2}{1/2}| = 3$. We need to find $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta}$. Substituting $	an 	heta = 3$,we get $	an 2	heta = \frac{2(3)}{1 - 3^2} = \frac{6}{1 - 9} = \frac{6}{-8} = -\frac{3}{4}$.

If the angle between the tangents drawn through the point $(-2, -1)$ to the parabola $y^2 = 4x$ is $\theta$,then $\tan 2\theta =$

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