This problem requires a blend of analytical geometry and integral calculus. We first need to find the equation of the tangent line to the given parabola from an external point and determine the point of tangency. Subsequently, we will use definite integration to calculate the area of the specified region bounded by this tangent line, the parabola, and the x-axis.

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1. Understanding the Parabola and the Problem Setup

Key Concept: A parabola of the form $y^2 = 4a(x-h)$ has its vertex at $(h,0)$ and opens towards the positive x-axis.

The given parabola is $P: y^2 = x-2$.

• Comparing this to $y^2 = 4a(x-h)$, we see that $4a=1 \implies a=1/4$, and $h=2$.
• Therefore, the vertex of the parabola is at $(2,0)$.
• The parabola opens to the right.
• The point A is given as $(-2,0)$. Notice that this point lies on the x-axis and is to the left of the parabola's vertex.
• The tangent line AB touches the parabola at point B in the first quadrant. This means the y-coordinate of B must be positive.
• Our goal is to find the coordinates of B, the equation of the line AB, and then the area of the region bounded by line AB, parabola P, and the x-axis.

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2. Finding the Equation of the Tangent Line AB and Point B

Key Concept: The equation of the tangent to the parabola $y^2 = 4aX$ at a point $(X_1, Y_1)$ is $Y Y_1 = 2a(X+X_1)$. For our parabola $y^2 = x-2$, we can consider $X = x-2$ and $Y=y$. So $Y^2 = X$. Here, $4a=1$, so $a=1/4$. Let the point of tangency be $B(x_B, y_B)$. The equation of the tangent at $B(x_B, y_B)$ to $y^2 = x-2$ is: $y y_B = \frac{1}{2}((x-2) + (x_B-2))$ $y y_B = \frac{1}{2}(x + x_B - 4) \quad \text{(Equation of Tangent)}$

Explanation: We use the standard tangent equation form. It's often more efficient than differentiation for parabolas, especially when dealing with external points. The "standard form" is derived from implicit differentiation and then simplifying.

Step 2.1: Use the external point A to find $x_B$. The tangent line passes through the point $A(-2,0)$. We substitute $x=-2$ and $y=0$ into the tangent equation: $0 \cdot y_B = \frac{1}{2}(-2 + x_B - 4)$ $0 = \frac{1}{2}(x_B - 6)$ $x_B - 6 = 0 \implies x_B = 6$

Explanation: Since the tangent line passes through A, the coordinates of A must satisfy the tangent's equation. This allows us to find one of the coordinates of the point of tangency, $x_B$.

Step 2.2: Find $y_B$ using the parabola's equation. Since $B(x_B, y_B)$ lies on the parabola $y^2 = x-2$, we substitute $x_B=6$ into the parabola's equation: $y_B^2 = 6-2$ $y_B^2 = 4$ $y_B = \pm 2$

Explanation: The point of tangency B must satisfy both the equation of the parabola and the equation of the tangent line. Using the parabola's equation helps us find the corresponding y-coordinate.

Step 2.3: Determine the correct $y_B$ and the coordinates of B. We are given that point B is in the first quadrant. Therefore, $y_B$ must be positive. So, $y_B = 2$. The point of tangency is $B(6,2)$.

Explanation: The problem statement specifies that B is in the first quadrant, which means both its x and y coordinates must be positive. This helps us disambiguate between the two possible y-values.

Step 2.4: Find the equation of the line AB. We have two points $A(-2,0)$ and $B(6,2)$. The slope of the line AB is $m = \frac{2-0}{6-(-2)} = \frac{2}{8} = \frac{1}{4}$. Using the point-slope form $y - y_1 = m(x - x_1)$ with $A(-2,0)$: $y - 0 = \frac{1}{4}(x - (-2))$ $y = \frac{1}{4}(x+2)$ $4y = x+2 \implies x - 4y + 2 = 0 \quad \text{(Equation of Line AB)}$

Explanation: With two points, we can uniquely determine the equation of a line. This line will serve as one of the boundaries for our area calculation.

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3. Visualizing and Setting Up the Area Calculation

Key Concept: The area of a region bounded by curves $y = f(x)$, $y = g(x)$ and vertical lines $x=a$, $x=b$ (where $f(x) \ge g(x)$ on $[a,b]$) is given by $\int_a^b (f(x) - g(x)) \, dx$.

Let's sketch the region:

• The parabola $