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 :
Show 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.
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.
