1. Key Concepts and Formulas Essential for Solving This Problem

To effectively solve this problem, we will leverage fundamental concepts from coordinate geometry and the theory of parabolas. Understanding these will make the solution clear and logical.

• Standard Parabola Equation and its Properties: For a parabola with its vertex at the origin and opening along the positive x-axis, its standard equation is $y^2 = 4Ax$.

  • Focus: The focus of this parabola is at the point $(A, 0)$.
  • Directrix: The equation of the directrix is $x = -A$.
  • Parametric Coordinates: Any point $P$ on this parabola can be represented parametrically as $P(At^2, 2At)$, where $t$ is a real parameter. This form is incredibly powerful as it allows us to represent a moving point using a single variable.

• Midpoint Formula: Given two points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$, the coordinates of their midpoint $M(h, k)$ are calculated as: $h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2}$ This formula helps us express the coordinates of the midpoint in terms of the coordinates of the two endpoints.

• General Parabola Equation and its Directrix: A parabola with its vertex at $(x_0, y_0)$ and axis parallel to the x-axis has the equation $(y - y_0)^2 = 4A_{new}(x - x_0)$. The directrix of such a parabola is given by the equation $x - x_0 = -A_{new}$, which simplifies to $x = x_0 - A_{new}$. This generalized form is crucial for identifying the properties of the locus we find.

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2. Problem Setup: Identifying Given Information and Our Goal

The problem asks us to find the locus of the mid-point of a line segment. This segment has two distinct endpoints:

1. The focus of a given parabola.
2. A moving point on that same parabola.

Let's break down the given information:

• The Given Parabola: The equation is $y^2 = 4ax$.

  • Explanation: This equation is in the standard form $y^2 = 4Ax$. By comparing $y^2 = 4ax$ with $y^2 = 4Ax$, we can clearly see that the parameter 'A' for this specific parabola is equal to $a$. This 'a' is a constant that defines the shape and position of the original parabola.

• The Focus of the Given Parabola (Fixed Point): Using the standard form $y^2 = 4Ax$, the focus is at $(A, 0)$. Since $A = a$ for our given parabola, its focus $S$ is at $(a, 0)$.

  • Explanation: This is a direct application of the focus formula. This point $S$ is fixed and does not change.

• The Moving Point on the Parabola: Let $P$ be a general point on the parabola $y^2 = 4ax$. Since $P$ is a "moving point," its coordinates will vary.

  • Explanation: To represent this moving point, we will use the parametric form. This is a powerful technique because it allows us to describe all points on the parabola using a single variable, which simplifies calculations when finding a locus.

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3. Expressing Coordinates of the Key Points

Now, let's write down the coordinates of the two points whose midpoint we are interested in.

• Focus $S$: $S(a, 0)$

  • Explanation: As derived in the previous section, by comparing $y^2 = 4ax$ with the standard form $y^2 = 4Ax$, we find that the focal length parameter for the given parabola is $a$.

• Moving Point $P$ on the Parabola: We use the parametric coordinates for a point on $y^2 = 4ax$. Since the parameter for this parabola is $a$, any point $P$ on it can be written as: $P(at^2, 2at)$

  • Explanation: This parametric form is chosen because it automatically satisfies the parabola's equation: $(2at)^2 = 4a(at^2) \implies 4a^2t^2 = 4a^2t^2$. As the parameter $t$ varies over all real numbers, the point $P$ traces out every point on the parabola. Using a single parameter $t$ is much more efficient for locus problems than using generic coordinates $(x_P, y_P)$ and the constraint $y_P^2 = 4ax_P$.

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4. Applying the Midpoint Formula

Let $M(h, k)$ be the midpoint of the line segment joining the focus $S(a, 0)$ and the moving point $P(at^2, 2at)$.

• Explanation: We denote the coordinates of the midpoint as $(h, k)$ because this is standard practice when finding a locus. Our ultimate goal is to find an equation that relates $h$ and $k$ but is independent of the parameter $t$. This equation will define the path (locus) of $M$.

Using the midpoint formula: $h = \frac{x_S + x_P}{2} = \frac{a + at^2}{2} \quad \cdots (1)$ $k = \frac{y_S + y_P}{2} = \frac{0 + 2at}{2} \quad \cdots (2)$

• Explanation: We are directly applying the midpoint formula to express the coordinates of $M$ in terms of the known coordinates of $S$ and the parametric coordinates of $P$. These two equations establish the relationship between the midpoint $(h, k)$ and the parameter $t$.

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5. Eliminating the Parameter 't' to Find the Locus

The next crucial step is to eliminate the parameter $t$ from equations (1) and (2). This process will yield a single equation solely in terms of $h$ and $k$, which represents the desired locus of $M$.

• Step 5a: Isolate $t$ from the simpler equation. Equation (2) is $k = \frac{2at}{2}$. Let's simplify and solve for $t$: $k = at$ From this, we can express $t$ in terms of $k$ and $a$: $t = \frac{k}{a} \quad \cdots (3)$

  • Explanation: We choose equation (2) to isolate $t$ because it is a linear equation in $t$, making the isolation straightforward and avoiding square roots at this stage.

• Step 5b: Substitute $t$ into the other equation. Before substituting $t$ into equation (1), it's useful to first calculate $t^2$ from equation (3): $t^2 = \left(\frac{k}{a}\right)^2 = \frac{k^2}{a^2}$ Now, substitute this expression for $t^2$ into equation (1): $h = \frac{a(1 + t^2)}{2}$ $h = \frac{a}{2}\left(1 + \frac{k^2}{a^2}\right)$

  • Explanation: This substitution is the mechanism to eliminate the parameter $t$. The resulting equation will then only contain $h$, $k$, and the constant $a$, representing the equation of the locus.

• Step 5c: Simplify the resulting equation. Let's simplify the equation to a more recognizable form: Multiply both sides by 2: $2h = a\left(1 + \frac{k^2}{a^2}\right)$ Distribute $a$ on the right side: $2h = a + \frac{ak^2}{a^2}$ $2h = a + \frac{k^2}{a}$

  • Explanation: Algebraic simplification makes the equation easier to manipulate and compare with standard conic section forms.

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6. Transforming to Standard Parabola Form and Identifying its Directrix

The equation $2h = a + \frac{k^2}{a}$ represents the locus of the midpoint $M(h, k)$. To identify its type and, specifically, its directrix, we need to rearrange it into a standard form of a parabola.

• Step 6a: Rearrange to match the standard parabola form. First, isolate the term involving $k^2$: $\frac{k^2}{a} = 2h - a$ Multiply by $a$: $k^2 = a(2h - a)$ To match the standard form $(Y - Y_0)^2 = 4A_{new}(X - X_0)$, we need to factor out $2a$ from the right side: $k^2 = 2a\left(h - \frac{a}{2}\right)$

  • Explanation: We manipulate the equation because we want to match it with the standard form of a parabola. This form allows us to easily identify the vertex, focal length, and directrix of the new parabola (the locus).

• Step 6b: Replace $(h, k)$ with $(x, y)$ to represent the locus. The locus of the midpoint $M(h, k)$ is conventionally represented by replacing $h$ with $x$ and $k$ with $y$: $y^2 = 2a\left(x - \frac{a}{2}\right)$

  • Explanation: This is a standard notation change. The equation in terms of $x$ and $y$ describes all points on the curve formed by the midpoints.

• Step 6c: Identify the parameters of this new parabola. Compare the equation $y^2 = 2a\left(x - \frac{a}{2}\right)$ with the standard form $(y - y_0)^2 = 4A_{new}(x - x_0)$. From this comparison, we can identify:

  • The vertex of the new parabola is $(x_0, y_0) = \left(\frac{a}{2}, 0\right)$.
  • The term $4A_{new}$ corresponds to $2a$. Therefore, the focal length of this new parabola, $A_{new}$, is: $4A_{new} = 2a \implies A_{new} = \frac{2a}{4} = \frac{a}{2}$
  • Explanation: By systematically comparing the derived locus equation with the general standard form, we extract its key properties. It's crucial to use a different symbol like $A_{new}$ (or $A'$) for the focal length of the new parabola to avoid confusion with the 'a' parameter of the original parabola.

• Step 6d: Determine the directrix of the new parabola. For a parabola of the form $(y - y_0)^2 = 4A_{new}(x - x_0)$, the equation of its directrix is given by $x - x_0 = -A_{new}$. Substitute the values we found: $x_0 = \frac{a}{2}$ and $A_{new} = \frac{a}{2}$. $x - \frac{a}{2} = - \frac{a}{2}$ $x = \frac{a}{2} - \frac{a}{2}$ $x = 0$

  • Explanation: We apply the standard directrix formula for a horizontally opening parabola. The directrix is a vertical line whose equation depends on the x-coordinate of the vertex ($x_0$) and the focal length ($A_{new}$).

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7. Tips for Success and Common Pitfalls

• Embrace Parametric Forms: For locus problems involving points on conic sections, parametric coordinates are almost always the most efficient approach. Invest time in learning the standard parametric forms for parabolas, ellipses, and hyperbolas.
• Systematic Parameter Elimination: The core of any locus problem is eliminating the parameter. Always isolate the parameter from the simplest equation (usually linear or involving only one variable) and substitute it into the more complex one.
• Master Standard Forms: Be thoroughly familiar with the standard equations of all conic sections and how to extract their key properties (vertex, focus, directrix, eccentricity, axes) from these forms.
• Distinguish Parameters: When dealing with multiple conic sections or transformations, it's good practice to use different symbols for similar parameters (e.g., 'a' for the original parabola's focal length and $A_{new}$ for the locus parabola's focal length). This prevents confusion.
• Coordinate Transformations: Remember that properties like the directrix $x = x_0 - A_{new}$ are relative to the original coordinate system. Carefully