1. Understanding the Parabola and Determining its Key Parameter 'a'

To begin, we need to understand the fundamental properties of the given parabola. The equation of a parabola dictates its shape, orientation, and the position of its focus and directrix.

Key Concept: The standard form of a parabola opening along the positive x-axis is $y^2 = 4ax$. In this form, 'a' is a crucial parameter. It defines the coordinates of the focus as $(a, 0)$ and the equation of the directrix as $x = -a$. This parameter 'a' is central to all calculations involving points, chords, and tangents of the parabola.

Step 1: Determine the value of 'a' for the given parabola.

The equation of the parabola provided in the problem is: $y^2 = 8x$

To find the value of 'a', we compare this equation with the standard form $y^2 = 4ax$. By equating the coefficients of $x$ from both equations: $4a = 8$ $a = \frac{8}{4}$ $a = 2$

Explanation: We determined $a=2$. This means the focus of our parabola is at $(2, 0)$ and its directrix is the line $x = -2$. This value of 'a' will be used consistently throughout the problem to define points and equations related to this specific parabola.

2. Representing Points on the Parabola Parametrically and Finding $t_A$

The problem gives us one endpoint of a focal chord, point A. To efficiently utilize properties of chords and tangents, especially focal chords, it is highly advantageous to represent points on the parabola using their parametric form.

Key Concept: Any point on the parabola $y^2 = 4ax$ can be expressed in parametric form as $P(at^2, 2at)$, where 't' is a real parameter. This representation simplifies many calculations, particularly when dealing with relationships between different points on the parabola, like those forming a chord or a tangent.

Step 2: Determine the parameter $t_A$ for point A.

The coordinates of point A are given as $A\left( \frac{1}{2}, -2 \right)$. Since A lies on the parabola, its coordinates must satisfy the parametric form $(at^2, 2at)$ with $a=2$. So, for point A, we can write: $x_A = at_A^2 \implies \frac{1}{2} = 2t_A^2$ $y_A = 2at_A \implies -2 = 2(2)t_A$

It's usually simpler and less prone to errors to solve for 't' using the $y$-coordinate equation first: $-2 = 4t_A$ $t_A = \frac{-2}{4}$ $t_A = -\frac{1}{2}$

Verification (Good Practice): To ensure accuracy, we can cross-check this value of $t_A$ using the $x$-coordinate equation: $x_A = 2t_A^2 = 2\left(-\frac{1}{2}\right)^2 = 2\left(\frac{1}{4}\right) = \frac{1}{2}$ This matches the given $x$-coordinate of A, confirming that $t_A = -\frac{1}{2}$ is correct.

Explanation: By converting the Cartesian coordinates of A into its parametric parameter $t_A$, we have prepared the point for easy application of the focal chord property in the next step. This is a standard and powerful technique in coordinate geometry.

3. Applying the Focal Chord Property to Find $t_B$

The problem explicitly states that AB is a focal chord. This is a crucial piece of information that establishes a direct relationship between the parameters of its endpoints, A and B.

Key Concept: If $P(at_1^2, 2at_1)$ and $Q(at_2^2, 2at_2)$ are the endpoints of a focal chord of the parabola $y^2 = 4ax$, then the product of their parameters is $t_1t_2 = -1$. This property implies that the parameters of the endpoints of any focal chord are negative reciprocals of each other.

Step 3: Determine the parameter $t_B$ for point B.

We have already found $t_A = -\frac{1}{2}$. Using the focal chord property $t_A t_B = -1$: $\left(-\frac{1}{2}\right) t_B = -1$ To solve for $t_B$, multiply both sides of the equation by -2: $t_B = (-1) \times (-2)$ $t_B = 2$

Explanation: This property is a significant shortcut. Instead of going through the tedious process of finding the focus, then the equation of the line AB passing through A and the focus, and finally intersecting this line with the parabola to find B, we can directly determine $t_B$ from $t_A$. This saves considerable time and reduces potential calculation errors.

Common Mistake Alert: A frequent error is to forget the negative sign in the focal chord property, mistakenly using $t_1t_2 = 1$ instead of $t_1t_2 = -1$. Always remember that the parameters are negative reciprocals.

4. Equation of the Tangent at Point B using its Parametric Form

With the parameter $t_B$ for point B and the value of 'a' for the parabola, we can now directly write the equation of the tangent at B.

Key Concept: The equation of the tangent to the parabola $y^2 = 4ax$ at a point $P(at^2, 2at)$ (i.e., at parameter 't') is given by the formula $ty = x + at^2$. This is the most efficient form to use when the parameter 't' of the point of tangency is known.

Step 4: Write the equation of the tangent at B.

We have $a=2$ and $t_B = 2$. Substitute these values into the parametric tangent equation $ty = x + at^2$: $(2)y = x + (2)(2)^2$ $2y = x + 2(4)$ $2y = x + 8$

To match the standard form of linear equations typically presented in options ($Ax + By + C = 0$), rearrange the equation: $x - 2y + 8 = 0$

Explanation: Using the parametric form of the tangent equation directly with $t_B$ and 'a' is significantly faster than first calculating the Cartesian coordinates of B ($B(at_B^2, 2at_B) = B(2(2^2), 2(2)(2)) = B(8, 8)$) and then applying the general tangent formula $yy_1 = 2a(x+x_1)$. The parametric approach streamlines the solution.

5. Conclusion and Final Answer

The equation of the tangent to the parabola at point B is $x - 2y + 8 = 0$.

Let's compare this result with the given options: (A) $2x – y – 24 = 0$ (B) $x – 2y + 8 = 0$ (C) $x + 2y + 8 = 0$ (D) $2x + y – 24 = 0$

Our derived equation perfectly matches option (B).

Summary and Key Takeaway: This problem beautifully illustrates the efficiency and elegance of using parametric representation in coordinate geometry, especially for parabolas. By following a structured approach:

1. Identifying the parameter 'a' from the parabola's equation.
2. Converting the given Cartesian coordinates of point A into its parametric form ($t_A$).
3. Leveraging the focal chord property ($t_A t_B = -1$) to find the parameter for the other end of the chord ($t_B$).
4. Directly applying the parametric form of the tangent equation ($ty = x + at^2$) using $t_B$ and 'a'. We can solve such problems quickly and accurately. Mastering these parametric forms and properties is crucial for JEE preparation.