A line passing through the point $ A(-2, 0) $, touches the parabola $ P: y^2 = x - 2 $ at the point $ B $ in the first quadrant. The area of the region bounded by the line $ AB $, parabola $ P $, and the x-axis is:
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To find the equation of a tangent line, use the point of tangency and solve for the slope using the discriminant.
Tangent The equation of the tangent line is expressed as \( y = m(x + 2) \). Upon substituting \( y^2 = x - 2 \) into this equation, we obtain \( (m(n + 2))^2 = n - 2 \). This simplifies to the quadratic equation \( m^2x^2 + (4m^2 - 1)x + (4m^2 + 2) = 0 \). For the line to be tangent, the discriminant must be zero, leading to \( D = 0 \Rightarrow (4m^2 - 1)^2 - 4m^2(4m^2 + 2) = 0 \). Solving for \( m \) yields \( m = \frac{1}{4} \). Substituting this value back into the equation for \( y \) gives \( y = \frac{1}{4}(n + 2) \). The point of tangency is determined to be \( (6, 2) \). The area is then calculated using the integral \( A = \int_0^2 \left( (y^2 + 2) - (4y - 2) \right) dy \). Evaluating this integral results in \( A = \frac{8}{3} \). Consequently, the correct option is \( (3) \).
