Question

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

• A. 6
• B. 9
• C. 10
• D. 4

Hint:

When calculating the area bounded by curves, it's crucial to choose the correct axis of integration. If the curves are given as \(x = f(y)\), integrating with respect to y is often much simpler. Always sketch the region to visualize the boundaries and determine which curve is on the right and which is on the left.

Answer 1

The Correct Option is B

Step 1: Find the point of tangency.
The ordinate (y-coordinate) is given as y=3. Substitute this into the parabola's equation to find the x-coordinate: \[ (3-2)^2 = x-1 \implies 1^2 = x-1 \implies x = 2 \] The point of tangency is P(2, 3). Step 2: Find the equation of the tangent line.
Differentiate the parabola's equation implicitly with respect to x: \[ 2(y-2)\frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{2(y-2)} \] At the point P(2, 3), the slope of the tangent is: \[ m = \frac{dy}{dx}\bigg|_{(2,3)} = \frac{1}{2(3-2)} = \frac{1}{2} \] The equation of the tangent line using point-slope form is \(y - y_1 = m(x - x_1)\): \[ y - 3 = \frac{1}{2}(x - 2) \implies 2y - 6 = x - 2 \implies x = 2y - 4 \] Step 3: Set up the integral for the area.
The region is bounded by the parabola \(x = (y-2)^2 + 1\), the tangent \(x = 2y - 4\), and the x-axis (\(y=0\)). It is most convenient to integrate with respect to y. The region extends from y=0 up to the point of tangency at y=3. The area between two curves \(x_{right}\) and \(x_{left}\) from \(y=c\) to \(y=d\) is given by \(\int_c^d (x_{right} - x_{left}) dy\). In the interval [0, 3], the parabola is to the right of the tangent line. \[ \text{Area} = \int_{0}^{3} (\text{Parabola} - \text{Tangent}) dy \] \[ \text{Area} = \int_{0}^{3} [((y-2)^2 + 1) - (2y - 4)] dy \] Step 4: Evaluate the integral.
\[ \text{Area} = \int_{0}^{3} (y^2 - 4y + 4 + 1 - 2y + 4) dy \] \[ \text{Area} = \int_{0}^{3} (y^2 - 6y + 9) dy \] This integrand is a perfect square: \[ \text{Area} = \int_{0}^{3} (y-3)^2 dy \] \[ \text{Area} = \left[ \frac{(y-3)^3}{3} \right]_0^3 \] \[ \text{Area} = \left( \frac{(3-3)^3}{3} \right) - \left( \frac{(0-3)^3}{3} \right) = 0 - \frac{-27}{3} = 9 \] The area of the region is 9 square units.