Key Concepts and Formulas

This problem requires a strong understanding of several core concepts from coordinate geometry, particularly related to parabolas, and basic number theory. Mastering these principles is essential for a systematic approach.

1. Equation of a Parabola: The standard form of a parabola symmetric about the x-axis, opening to the right, is $y^2 = 4Ax$. Here, $A$ represents the focal length. The given parabola is $y^2 = 2ax$. By comparing these forms, we can identify $4A = 2a$, which implies $A = \frac{a}{2}$. This parameter $A$ is crucial for constructing the tangent equation.

2. Equation of the Tangent to a Parabola: For a parabola $y^2 = 4Ax$, the equation of the tangent line at a point $(x_1, y_1)$ lying on the parabola is given by $yy_1 = 2A(x+x_1)$. This formula can be derived using calculus (finding the derivative $\frac{dy}{dx}$ to get the slope and then using the point-slope form) or by a direct method of replacing $y^2$ with $yy_1$ and $x$ with $\frac{x+x_1}{2}$.

3. Area of a Triangle: Given the coordinates of the three vertices of a triangle, say $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, the area can be calculated using the determinant formula $\frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$. However, for triangles where one side lies on a coordinate axis (like the x-axis in this problem) and the opposite vertex is known, a simpler formula is $\frac{1}{2} \times \text{base} \times \text{height}$.

4. Natural Numbers ($\mathbb{N}$): The problem states that $a, b, c \in \mathbb{N}$. This is a critical constraint, meaning $a, b, c$ must be positive integers ($1, 2, 3, \ldots$). This will help us filter out non-integer or non-positive solutions later.

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Understanding the Problem and Initial Setup

We are given a parabola $y^2 = 2ax$. A specific point $P(b, c)$ with $b, c \in \mathbb{N}$ lies on this parabola. A tangent line is drawn to the parabola at $P$. This tangent line, along with the vertical line $x=b$ and the x-axis ($y=0$), forms a triangle. The area of this triangle is given as $16$ square units. Our objective is to find all possible natural number values for $a$ (which form the set $S$) and then calculate the sum of these values.

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Step 1: Condition for Point P on the Parabola

The first piece of information is that the point $P(b, c)$ lies on the parabola $y^2 = 2ax$. This means its coordinates must satisfy the parabola's equation. Why this step? This establishes the first fundamental relationship between the parameters $a, b,$ and $c$. This equation is crucial as it connects the point of tangency to the specific parabola.

Substituting $x=b$ and $y=c$ into the parabola's equation: $c^2 = 2ab \quad \ldots (1)$ Self-check: Since $a, b, c \in \mathbb{N}$, we know $a \ge 1$, $b \ge 1$, and $c \ge 1$. This ensures that $2ab$ is a positive integer, and thus $c^2$ is also a positive integer, which is consistent with $c \in \mathbb{N}$.

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Step 2: Equation of the Tangent Line

Next, we need to find the equation of the tangent to the parabola $y^2 = 2ax$ at the point $P(b, c)$. Why this step? The tangent line is one of the three lines forming the triangle. To find the vertices of the triangle, we need its equation.

Recall the general formula for the tangent to $y^2 = 4Ax$ at $(x_1, y_1)$ is $yy_1 = 2A(x+x_1)$. For our parabola, $y^2 = 2ax$, we compare it with $y^2 = 4Ax$ to find $4A = 2a$, which implies $2A = a$. The point of tangency is $(x_1, y_1) = (b, c)$. Substituting these values into the tangent formula: $yc = a(x+b) \quad \ldots (2)$ This is the equation of the tangent line.

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Step 3: Identifying the Vertices of the Triangle

The triangle is formed by three lines:

1. Tangent Line ($L_T$): $yc = a(x+b)$
2. Vertical Line ($L_V$): $x=b$
3. Horizontal Line ($L_H$): $y=0$ (the x-axis)

Why this step? To calculate the area of the triangle, we first need to know its vertices. We find these by determining the intersection points of these three lines.

Let's find the intersection points (vertices):

• Vertex 1 (P): This is the intersection of the tangent line ($L_T$) and the vertical line ($L_V$). Since the tangent is drawn at the point $P(b, c)$, and the line $x=b$ is a vertical line passing through $P$, their intersection is simply $P(b, c)$ itself. So, $V_1 = P(b, c)$.

• Vertex 2 (Q): This is the intersection of the tangent line ($L_T$) and the x-axis ($L_H$, where $y=0$). Substitute $y=0$ into the tangent equation $yc = a(x+b)$: $(0)c = a(x+b)$ $0 = a(x+b)$ Since $a \in \mathbb{N}$, $a$ is a positive integer and therefore cannot be zero. Thus, we must have $x+b=0$, which implies $x=-b$. So, $V_2 = Q(-b, 0)$.

• Vertex 3 (R): This is the intersection of the vertical line ($L_V$, $x=b$) and the x-axis ($L_H$, $y=0$). The coordinates are directly given by these two equations: $x=b$ and $y=0$. So, $V_3 = R(b, 0)$.

The vertices of the triangle are $P(b, c)$, $Q(-b, 0)$, and $R(b, 0)$. Tip: It's highly recommended to quickly sketch these points. You'll notice that $Q$ and $R$ lie on the x-axis, and $P$ has the same x-coordinate as $R$. This structure indicates that the triangle is a right-angled triangle, with the right angle at $R(b,0)$.

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Step 4: Calculating the Area of the Triangle

We have the vertices $P(b, c)$, $Q(-b, 0)$, and $R(b, 0)$. Why this step? The problem provides the area (16 units$^2$). Using this information, we can form another equation relating $b$ and $c$, which will be crucial for finding possible values.

The segment $QR$ lies on the x-axis, making it a convenient base for the triangle.

• Length of the base $QR$: The distance between $Q(-b, 0)$ and $R(b, 0)$ is $|b - (-b)| = |2b|$. Since $b \in \mathbb{N}$, $b$ is a positive integer, so $2b > 0$. Thus, Base $= 2b$.
• Height of the triangle: The height is the perpendicular distance from vertex $P(b, c)$ to the base $QR$ (which lies on the x-axis). This distance is simply the absolute value of the y-coordinate of $P$, which is $|c|$. Since $c \in \mathbb{N}$, $c$ is a positive integer, so $c > 0$. Thus, Height $= c$.

Now, we can calculate the area of the triangle: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (2b) \times c = bc$ The problem states that the area of the triangle is $16$ unit$^2$. Therefore, we have our second crucial relationship: $bc = 16 \quad \ldots (3)$

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Step 5: Finding Possible Natural Number Values for b and c

From equation (3), $bc=16$. We are given that $b, c \in \mathbb{N}$ (natural numbers), meaning they must be positive integers. Why this step? By listing all possible pairs of $(b, c)$ that satisfy $bc=16$ and the natural number constraint, we create a set of discrete possibilities that we can test in the next step to find $a$.

We need to list all pairs of positive integers $(b, c)$ whose product is 16. The factors of 16 are 1, 2, 4, 8, 16. The possible pairs $(b, c)$ are:

1. $(b, c) = (1, 16)$
2. $(b, c) = (2, 8)$
3. $(b, c) = (4, 4)$
4. $(b, c) = (8, 2)$
5. $(b, c) = (16, 1)$

These are all the possible combinations for $b$ and $c$ that satisfy both the area condition and the natural number constraint.

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Step 6: Determining Valid Natural Number Values for 'a'

Now we use equation (1), $c^2 = 2ab$, along with the possible pairs of $(b, c)$ found in Step 5, to find the corresponding values of $a$. Why this step? This is the final filter. For each valid $(b,c)$ pair, we calculate $a$ and check if $a$ itself is a natural number, as specified in the problem statement ($a \in \mathbb{N}$).

From $c^2 = 2ab$, we can express $a$ as: $a = \frac{c^2}{2b}$

Let's test each pair $(b, c)$:

1. For $(b, c) = (1, 16)$: $a = \frac{16^2}{2 \times 1} = \frac{256}{2} = 128$ Since $128 \in \mathbb{N}$, $a=128$ is a valid value.

2. For $(b, c) = (2, 8)$: $a = \frac{8^2}{2 \times 2} = \frac{64}{4} = 16$ Since $16 \in \mathbb{N}$, $a=16$ is a valid value.

3. For $(b, c) = (4, 4)$: $a = \frac{4^2}{2 \times 4} = \frac{16}{8} = 2$ Since $2 \in \mathbb{N}$, $a=2$ is a valid value.

4. For $(b, c) = (8, 2)$: $a = \frac{2^2}{2 \times 8} = \frac{4}{16} = \frac{1}{4}$ Since $\frac{1}{4}$ is not a natural number ($a \notin \mathbb{N}$), this value of $a$ is not valid.

5. For $(b, c) = (16, 1)$: $a = \frac{1^2}{2 \times 16} = \frac{1}{32}$ Since $\frac{1}{32}$ is not a natural number ($a \notin \mathbb{N}$), this value of $a$ is not valid.

The set $S$ of all valid natural number values for $a$ is $S = \{128, 16, 2\}$.

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Step 7: Summing the Values in S

The question asks for the sum of all $a \in S$. Why this step? This is the final calculation required by the problem statement.

$\sum_{a \in S} a = 128 + 16 + 2 = 146$

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Final Answer Check and Summary

Let's quickly re-verify one case, e.g., $a=2$. If $a=2$, the parabola is $y^2 = 4x$. For $(b,c)=(4,4)$, $P(4,4)$ lies on $y^2=4x$ because $4^2 = 16 = 4(4)$. This is correct ($c^2=2ab \implies 4^2 = 2(2)(4) \implies 16 = 16$). The tangent at $(4,4)$ to $y^2=4x