Question

The area of the region bounded by the parabola \( (y - 2)^2 = (x - 1) \), the tangent to the parabola at the point \( (2, 3) \), and the X-axis is:

• A. 3
• B. 6
• C. 9
• D. 12

Hint:

To find the area between a curve and a line, calculate the area under each function over the given range and subtract the areas.

Answer 1

The Correct Option is C

Calculation of Area under a Parabola and above a Tangent Line 
The given parabola equation is: \[ y^2 - 4y - x + 5 = 0. \] The equation of the tangent at the point \( (2, 3) \) is derived to be: \[ 3y - 2(y+3) - \frac{x+2}{2} + 5 = 0. \] Simplifying the equation, we get: \[ 2y - x - 4 = 0. \] The area, \( A \), required is the region bounded by this tangent line and the parabola between their points of intersection. We can calculate this area by integrating between the bounds set by these intersections. The correct limits of integration and the functions need first to be determined by setting \( 2y - x - 4 = 0 \) equal to \( y = \frac{x+4}{2} \) and substituting into the parabola's equation. Solving the resulting equations, we find: \[ y^2 - 4y - \left(\frac{y-2}{2}\right) + 5 = 0. \] This integrates to: \[ A = \int_0^3 (x_2 - x_1) \, dy = \int_0^3 \left[(y-2)^2 + 1 - \left(2y - 4\right)\right] \, dy, \] which simplifies to: \[ A = \int_0^3 \left((3-y)^2\right) \, dy. \] Calculating this integral, we find: \[ A = \left[ -\frac{(3-y)^3}{3} \right]_0^3 = 9. \] This calculation concludes that the area of the region specified is 9 square units. Thus, the correct answer is Option C.