Angle Subtended by Parabola Focal Chord at Vertex

To find $\tan \theta$, where $\theta$ is the angle subtended by the focal chord $PQ$ at the vertex of the parabola $y^2 = 4ax$, we follow these steps:

Step 1: Parametric representation of points $P$ and $Q$

Let the parametric coordinates of points $P$ and $Q$ on the parabola $y^2 = 4ax$ be: $P = (at_1^2, 2at_1) \quad \text{and} \quad Q = (at_2^2, 2at_2)$

Since $PQ$ is a focal chord of the parabola, the parameters $t_1$ and $t_2$ satisfy: $t_1 t_2 = -1$

Step 2: Intersection of tangents

The point of intersection $T$ of the tangents drawn at $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$ is given by: $T = \big(at_1 t_2, \, a(t_1 + t_2)\big)$

Substituting $t_1 t_2 = -1$: $T = \big(-a, \, a(t_1 + t_2)\big)$

Step 3: Using the line condition

It is given that $T$ lies on the line $y = 2x + a$. Substituting the coordinates of $T$ into this equation: $a(t_1 + t_2) = 2(-a) + a$ $a(t_1 + t_2) = -a$

Since $a > 0$, we divide by $a$ to get: $t_1 + t_2 = -1$

Now, we compute $|t_1 - t_2|$ using the identity $(t_1 - t_2)^2 = (t_1 + t_2)^2 - 4t_1 t_2$: $(t_1 - t_2)^2 = (-1)^2 - 4(-1) = 1 + 4 = 5$ $|t_1 - t_2| = \sqrt{5}$

Step 4: Finding the angle subtended at the vertex

The vertex of the parabola $y^2 = 4ax$ is at the origin $V(0, 0)$.

The slope $m_1$ of the line $VP$ and the slope $m_2$ of the line $VQ$ are: $m_1 = \frac{2at_1 - 0}{at_1^2 - 0} = \frac{2}{t_1}$ $m_2 = \frac{2at_2 - 0}{at_2^2 - 0} = \frac{2}{t_2}$

The product of the slopes is: $m_1 m_2 = \left(\frac{2}{t_1}\right)\left(\frac{2}{t_2}\right) = \frac{4}{t_1 t_2} = \frac{4}{-1} = -4$

The magnitude of the difference between the slopes is: $|m_1 - m_2| = \left| \frac{2}{t_1} - \frac{2}{t_2} \right| = \frac{2|t_2 - t_1|}{|t_1 t_2|} = \frac{2\sqrt{5}}{1} = 2\sqrt{5}$

The angle $\theta$ subtended at the vertex satisfies the tangent angle formula: $\tan \theta = \frac{m_1 - m_2}{1 + m_1 m_2}$

Since $1 + m_1 m_2 = 1 - 4 = -3 < 0$, the angle $\theta$ is obtuse, making $\tan \theta$ negative: $\tan \theta = \frac{|m_1 - m_2|}{1 + m_1 m_2} = \frac{2\sqrt{5}}{-3} = -\frac{2}{3}\sqrt{5}$

Thus, the correct option is D.