Question

The area of the figure bounded by the parabola \((y-2)^2 = x-1\), the tangent to it at the point with the ordinate 3 and the \(x\)-axis is

• A. 3
• B. 6
• C. 9
• D. None of these

Hint:

Integrate with respect to \(y\) since parabola is sideways.

Answer 1

The Correct Option is C

To find the area of the region bounded by the parabola \((y-2)^2 = x-1\), the tangent to it at the point with the ordinate 3, and the \(x\)-axis, we proceed as follows:

1. Identify the given curve: The equation \((y-2)^2 = x-1\) is a sideways opening parabola with vertex at \((1, 2)\).
2. Find the point of tangency: We need the tangent at the point where \( y = 3 \). Substitute \( y = 3 \) into the equation of the parabola to find \( x \):
   \[ (3-2)^2 = x-1 \] \[ 1 = x-1 \] \[ x = 2 \]
   Thus, the point of tangency is \( (2, 3) \).
3. Find the slope of the tangent: Differentiate the parabola equation with respect to \( y \):
   \[ \frac{d}{dy}[(y-2)^2] = \frac{d}{dy}[x-1] \] \[ 2(y-2)\frac{dy}{dy} = \frac{dx}{dy} \] \[ \frac{dx}{dy} = 2(y-2) \]
   At \( y = 3 \), \(\frac{dx}{dy} = 2(3-2) = 2\).
4. Equation of the tangent line: With the point-slope form of the line equation, \((y_0 - y_1) = m(x - x_1)\), where the slope \( m = 2 \) and the point is \((2, 3)\):
   \[ y - 3 = 2(x - 2) \] \[ y = 2x - 4 + 3 \] \[ y = 2x - 1 \]
5. Find intersections with x-axis: To find intersections of the tangent with the \(x\)-axis, put \( y = 0 \) in the tangent equation:
   \[ 0 = 2x - 1 \] \[ 2x = 1 \] \[ x = \frac{1}{2} \]
   This point of intersection is \(\left( \frac{1}{2}, 0 \right)\).
6. Area calculation: The x-coordinates of the bounded region are from \(\frac{1}{2}\) to 2. The area under the tangent line and above the x-axis forms a triangle with vertices \(\left( \frac{1}{2}, 0 \right)\), \((2, 3)\), and \((2, 0)\).
7. Area of the triangle: Using the formula for the area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
   \[\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|\] \[ \text{Area} = \frac{1}{2} \left| \frac{1}{2} (3 - 0) + 2(0 - 0) + 2(0 - 3) \right| \] \[ = \frac{1}{2} \left| \frac{3}{2} - 6 \right| \] \[ = \frac{1}{2} \times \left| -\frac{9}{2} \right| \] \[ = \frac{9}{2} \]
   The area is \[9 \, \text{square units}\].

Thus, the area of the region bounded by the parabola, the tangent line, and the \(x\)-axis is 9 square units.