<p>A focal chord of parabola \(y^2 = 4ax\) is of length 4<i>a</i>. The angle subtended by it at the vertex of the parabola is <i>θ</i>, then \(|\tan θ|\) is equal to:</p>
Step-by-Step Solution
Key Concept: Use the focal chord property that for endpoints (at₁², 2at₁) and (at₂², 2at₂) on parabola y² = 4ax, the product t₁t₂ = -1. Combined with the chord length formula and angle calculation at vertex, we can find |tan θ|.
Step 1: Set up focal chord endpoints.
The parabola is $y^2 = 4ax$. Its vertex is $V(0,0)$.
Let the endpoints of the focal chord be $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$.
For a focal chord, the parameters $t_1$ and $t_2$ satisfy the condition $t_1t_2 = -1$.
Step 2: Use focal chord length formula.
The length of a focal chord is given by $L = a(t_1 - t_2)^2$.
Using the condition $t_2 = -1/t_1$, the length formula becomes:
$$L = a\left(t_1 - \left(-\frac{1}{t_1}\right)\right)^2 = a\left(t_1 + \frac{1}{t_1}\right)^2$$
Given that the length of the focal chord is $4a$, we have:
$$a\left(t_1 + \frac{1}{t_1}\right)^2 = 4a$$
$$\left(t_1 + \frac{1}{t_1}\right)^2 = 4$$
$$t_1 + \frac{1}{t_1} = \pm 2$$
Step 3: Solve for $t_1$ and $t_2$.
Case 1: $t_1 + \frac{1}{t_1} = 2$
Multiplying by $t_1$ (assuming $t_1 \neq 0$):
$$t_1^2 + 1 = 2t_1$$
$$t_1^2 - 2t_1 + 1 = 0$$
$$(t_1 - 1)^2 = 0$$
$$t_1 = 1$$
Then, $t_2 = -1/t_1 = -1/1 = -1$.
Case 2: $t_1 + \frac{1}{t_1} = -2$
Multiplying by $t_1$:
$$t_1^2 + 1 = -2t_1$$
$$t_1^2 + 2t_1 + 1 = 0$$
$$(t_1 + 1)^2 = 0$$
$$t_1 = -1$$
Then, $t_2 = -1/t_1 = -1/(-1) = 1$.
Both cases yield the same pair of parameters $\{1, -1\}$ for the endpoints of the focal chord.
Step 4: Find coordinates of endpoints.
Using $t_1 = 1$ and $t_2 = -1$:
The coordinates of point $P$ are $(a(1)^2, 2a(1)) = (a, 2a)$.
The coordinates of point $Q$ are $(a(-1)^2, 2a(-1)) = (a, -2a)$.
Step 5: Calculate the angle at the vertex.
The vertex of the parabola is $V(0,0)$. The angle $\theta$ is subtended by the chord $PQ$ at the vertex $V$, which means it is the angle between the line segments $VP$ and $VQ$.
The slope of the line $VP$ is $m_1 = \frac{2a - 0}{a - 0} = 2$.
The slope of the line $VQ$ is $m_2 = \frac{-2a - 0}{a - 0} = -2$.
The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula:
$$\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|$$
Substituting the slopes $m_1 = 2$ and $m_2 = -2$:
$$\tan \theta = \left|\frac{2 - (-2)}{1 + (2)(-2)}\right|$$
$$\tan \theta = \left|\frac{4}{1 - 4}\right|$$
$$\tan \theta = \left|\frac{4}{-3}\right|$$
$$\tan \theta = \frac{4}{3}$$
Therefore, $|\tan \theta| = \frac{4}{3}$.
Correct Answer: p