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(A) The equation of any tangent to the parabola $y^2 = 4x$ is given by $y = mx + \frac{1}{m}$.Since the tangent passes through the point $(-2, -1)$,we

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$3$

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(A) The equation of any tangent to the parabola $y^2 = 4x$ is given by $y = mx + \frac{1}{m}$.Since the tangent passes through the point $(-2, -1)$,we have:$-1 = -2m + \frac{1}{m}$$-m = -2m^2 + 1$$2m^2 - m - 1 = 0$Let the slopes of the two tangents be $m_1$ and $m_2$. From the quadratic equation,we have:$m_1 + m_2 = \frac{1}{2}$$m_1 m_2 = -\frac{1}{2}$The angle $\alpha$ between the two tangents is given by:$	an \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$Using the identity $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$:$(m_1 - m_2)^2 = (\frac{1}{2})^2 - 4(-\frac{1}{2}) = \frac{1}{4} + 2 = \frac{9}{4}$$|m_1 - m_2| = \frac{3}{2}$Substituting the values:$	an \alpha = \left| \frac{3/2}{1 - 1/2} ight| = \left| \frac{3/2}{1/2} ight| = 3$

Find the angle $\alpha$ between the tangents drawn from the point $(-2, -1)$ to the parabola $y^2 = 4x$. What is the value of $\tan \alpha$?

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