This problem is a comprehensive test of your understanding of parabola properties and basic coordinate geometry, specifically involving tangents, normals, directrices, and the properties of a square. Let's break down the solution step-by-step to ensure clarity and a thorough understanding.

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1. Understanding the Parabola and its Directrix

• Key Concept: The standard equation of a parabola opening to the right is $y^2 = 4ax$. For such a parabola, its directrix is the vertical line $x = -a$.
• Why this step is important: To accurately determine the directrix and subsequent equations for tangent and normal, we first need to identify the parameter 'a' from the given parabola equation.
• Working: We are given the equation of the parabola as $y^2 = 8x$. Comparing this with the standard form $y^2 = 4ax$: $4a = 8$ $a = 2$ Now, we can find the equation of the directrix. For $a=2$, the directrix is: $x = -a \implies x = -2$ This is a vertical line.

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2. Equation of the Tangent at Point P and Finding Point A

• Key Concept: The equation of the tangent to the parabola $y^2 = 4ax$ at a point $(x_1, y_1)$ on the parabola is given by $yy_1 = 2a(x + x_1)$.
• Why this step is important: We need the equation of the tangent line to find its intersection point with the directrix, which is defined as point A.
• Working: The given point P is $(2, -4)$. We have $x_1 = 2$, $y_1 = -4$, and $a = 2$ from the previous step. Substitute these values into the tangent formula: $y(-4) = 2(2)(x + 2)$ $-4y = 4(x + 2)$ Divide both sides by 4: $-y = x + 2$ Rearrange to the standard form $Ax + By + C = 0$: $x + y + 2 = 0 \quad \text{(Equation of Tangent)}$ To find point A, we intersect this tangent line with the directrix $x = -2$: Substitute $x = -2$ into the tangent equation: $(-2) + y + 2 = 0$ $y = 0$ Therefore, point A is $(-2, 0)$.

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3. Equation of the Normal at Point P and Finding Point B

• Key Concept: The normal to a curve at a point is perpendicular to the tangent at that point. If the slope of the tangent is $m_T$, then the slope of the normal $m_N$ is its negative reciprocal, i.e., $m_N = -\frac{1}{m_T}$. The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
• Why this step is important: Similar to finding point A, we need the equation of the normal line to find its intersection point with the directrix, which is defined as point B.
• Working: From the tangent equation $x + y + 2 = 0$, we can find its slope. Rewriting it as $y = -x - 2$, the slope of the tangent $m_T = -1$. The slope of the normal $m_N$ will be the negative reciprocal of $m_T$: $m_N = -\frac{1}{(-1)} = 1$ Now, use the point-slope form for the normal line, passing through $P(2, -4)$ with slope $m_N = 1$: $y - (-4) = 1(x - 2)$ $y + 4 = x - 2$ Rearrange to the standard form $Ax + By + C = 0$: $x - y - 6 = 0 \quad \text{(Equation of Normal)}$ To find point B, we intersect this normal line with the directrix $x = -2$: Substitute $x = -2$ into the normal equation: $(-2) - y - 6 = 0$ $-y - 8 = 0$ $y = -8$ Therefore, point B is $(-2, -8)$.

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4. Utilizing Square Properties to Find Point Q(a, b)

• Key Concept: In a square (or any parallelogram), the diagonals bisect each other. This means the midpoint of one diagonal is the same as the midpoint of the other diagonal. The midpoint formula for two points $(x_1, y_1)$ and $(x_2, y_2)$ is $M = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
• Why this step is important: This property provides the most efficient way to find the coordinates of the fourth vertex Q, given the other three vertices A, B, and P, and the fact that AQBP forms a square. Using side lengths or perpendicularity conditions would involve more complex calculations.
• Working: We are given that AQBP is a square. The vertices are $A(-2, 0)$, $Q(a, b)$, $B(-2, -8)$, and $P(2, -4)$. The diagonals of the square are AB and PQ. First, calculate the midpoint of diagonal AB: $M_{AB} = \left(\frac{-2 + (-2)}{2}, \frac{0 + (-8)}{2}\right) = \left(\frac{-4}{2}, \frac{-8}{2}\right) = (-2, -4)$ Next, calculate the midpoint of diagonal PQ: $M_{PQ} = \left(\frac{a + 2}{2}, \frac{b + (-4)}{2}\right) = \left(\frac{a + 2}{2}, \frac{b - 4}{2}\right)$ Since the diagonals bisect each other, $M_{AB}$ must be equal to $M_{PQ}$: $(-2, -4) = \left(\frac{a + 2}{2}, \frac{b - 4}{2}\right)$ Equating the x-coordinates: $-2 = \frac{a + 2}{2}$ $-4 = a + 2$ $a = -6$ Equating the y-coordinates: $-4 = \frac{b - 4}{2}$ $-8 = b - 4$ $b = -4$ So, the coordinates of point Q are $(-6, -4)$.

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5. Calculate the Final Expression $2a + b$

• Key Concept: Simple substitution into an algebraic expression.
• Why this step is important: This is the final step to answer the specific question asked in the problem.
• Working: We found $a = -6$ and $b = -4$. Substitute these values into the expression $2a + b$: $2a + b = 2(-6) + (-4)$ $2a + b = -12 - 4$ $2a + b = -16$

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Tips and Common Mistakes

• Verify the Point: Always quickly check if the given point P lies on the parabola. Substitute its coordinates into the parabola's equation: $(-4)^2 = 16$ and $8(2) = 16$. It matches, so P(2, -4) is indeed on the parabola.
• Directrix Sign: Remember that for $y^2 = 4ax$, the directrix is $x = -a$. For $x^2 = 4ay$, it's $y = -a$. Be careful with signs.
• Slope Calculation: Double-check your tangent and normal slope calculations. A common error is forgetting the negative reciprocal for the normal's slope.
• Quadrilateral Vertex Order: When a quadrilateral is named (e.g., AQBP), the vertices are usually listed in order around the perimeter. This means AB and PQ are the diagonals. If it were APBQ, then AB and PQ would be sides, and AP and BQ would be diagonals. This distinction is crucial for applying the midpoint property correctly.
• Visualization: A quick sketch of the parabola, directrix, and the points A, B, P, and Q can help you visualize the square and catch any obvious coordinate errors. For instance, A and B are on the directrix $x=-2$, so they should form a vertical line segment. P(2, -4) is to the right. Q should be to the left of A and B to complete the square.

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Summary and Key Takeaway

This problem beautifully integrates concepts from coordinate geometry of parabolas with the geometric properties of a square. The key steps involved:

1. Identifying the parabola's parameters and directrix.
2. Calculating the equations of the tangent and normal at a given point.
3. Finding the intersection points of these lines with the directrix.
4. Crucially, using the property that the diagonals of a square bisect each other to find the missing vertex.
5. Performing a final algebraic calculation.

Mastering these fundamental techniques for parabolas and applying basic geometric properties efficiently is vital for success in JEE Mathematics.

The final answer is $\boxed{-16}$.