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

To solve this problem, we need to determine the area of the region bounded by the given parabola, its tangent at a specified point, and the x-axis.

The equation of the given parabola is: (y - 2)^2 = (x - 1) .

Step 1: Determine the equation of the tangent

For a point on the parabola where the ordinate is 3, substitute \( y = 3 \) into the parabola equation:

\((3 - 2)^2 = (x - 1)\)

Solving,

1 = (x - 1)

So, \(x = 2\). The point of tangency is \((2, 3)\).

To find the tangent, differentiate the parabola equation with respect to \(x\):

\(\frac{d}{dx}[(y - 2)^2] = \frac{d}{dx}(x - 1)\)

Using chain rule,

2(y - 2) \cdot \frac{dy}{dx} = 1

At the point \((2, 3)\),

2(3 - 2) \cdot \frac{dy}{dx} = 1 \Rightarrow 2 \cdot \frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = \frac{1}{2}

The slope of the tangent is \(\frac{1}{2}\). Therefore, the equation of the tangent is given by the point-slope form:

y - 3 = \frac{1}{2}(x - 2)

Simplifying,

y = \frac{1}{2}x + 2

Step 2: Determine the points of intersection

Find the x-intercepts of the parabola (when \( y = 0\)) and x-intercepts of the tangent (when \(y = 0\)).

For the parabola:

(0 - 2)^2 = (x - 1)

Solving gives:

4 = x - 1, \Rightarrow x = 5

For the tangent line:

0 = \frac{1}{2}x + 2 \Rightarrow \frac{1}{2}x = -2 \Rightarrow x = -4

Step 3: Calculate Area

The region bounded is between \( x = -4 \) and \( x = 2 \) under the tangent, and \( x = 2 \) and \( x = 5 \) under the parabola.

Area under the tangent from \( x = -4 \) to \( x = 2 \):

\int_{-4}^{2} \left(\frac{1}{2}x + 2\right) dx

Computing this gives,

= \left[ \frac{1}{4}x^2 + 2x \right]_{-4}^{2} = \left(1+[4\right) - \left(4 + [-8]\right) = 9

Conclusion

The area of the bounded region is 9 square units.

The correct answer is: Option 9.