JEE Main 2009

LEVELJEE Main

Animated Solution for Mathematics - Definite Integration: The area of the region bounded by the parabola $(y - 2)^{2} = x - 1$ , the tangent of the parabola at the point (2, 3) and the x-axis is:

Q: Why do we integrate with respect to $y$ instead of $x$?

Great question! If you integrate with respect to $x$, you would have to split the region into two parts: one part bounded by the parabola's lower and upper branches, and another part bounded by the tangent line. By integrating with respect to $y$, the boundaries are simply $x = (y-2)^2 + 1$ on the right and $x = 2y - 4$ on the left, allowing us to compute the area in one clean, elegant step.

Q: How did we get the equation of the tangent line as $x = 2y - 4$?

We used the point-slope form of a line, $y - y_1 = m(x - x_1)$. We found the slope $m$ by differentiating the parabola's equation implicitly, giving us $m = \frac{1}{2}$ at the point $(2, 3)$. Substituting these into the formula $y - 3 = \frac{1}{2}(x - 2)$ and rearranging for $x$ gives $x = 2y - 4$.

Select Answer:

Visualized Solution

Visualize the Parabola $(y - 2)^{2} = x - 1$

Given equation of the parabola: $(y - 2)^{2} = x - 1$
This is a horizontal parabola opening to the right.
The vertex of this parabola is at $(1, 2)$ .

Differentiate to Find the Slope Function

To find the slope of the tangent, we differentiate the equation with respect to $x$ .
Differentiating $(y - 2)^{2} = x - 1$ implicitly:
$2 (y - 2) \frac{d y}{d x} = 1$

Calculate Slope at Point $(2, 3)$

We are given the point of tangency: $(2, 3)$
Substitute $y = 3$ into the derivative equation:
$2 (3 - 2) \frac{d y}{d x} = 1 ⟹ 2 (1) \frac{d y}{d x} = 1$
Slope $m = \frac{d y}{d x} = \frac{1}{2}$

Equation of the Tangent Line

Using the point-slope form: $y - y_{1} = m (x - x_{1})$
Substitute $(x_{1}, y_{1}) = (2, 3)$ and $m = \frac{1}{2}$ :
$y - 3 = \frac{1}{2} (x - 2)$
Simplifying: $2 y - 6 = x - 2 ⟹ x = 2 y - 4$

Identify the Bounded Region and Limits

The region is bounded by:
1. The parabola: $x = (y - 2)^{2} + 1$
2. The tangent line: $x = 2 y - 4$
3. The $x$ -axis: $y = 0$
The region extends vertically from $y = 0$ to $y = 3$ .

Set up the Integral with respect to $y$

Area formula: $A = \int_{y_{1}}^{y_{2}} (x_{right} - x_{left}) d y$
Here, $x_{right} = (y - 2)^{2} + 1$ (Parabola)
And $x_{left} = 2 y - 4$ (Tangent)
$A = \int_{0}^{3} [((y - 2)^{2} + 1) - (2 y - 4)] d y$

Simplify the Algebraic Expression

Expand the terms inside the integrand:
$((y - 2)^{2} + 1) - (2 y - 4) = (y^{2} - 4 y + 4 + 1) - 2 y + 4$
$= y^{2} - 6 y + 9$
Notice that this is a perfect square: $(y - 3)^{2}$

Integrate the Simplified Function

We need to evaluate: $A = \int_{0}^{3} (y - 3)^{2} d y$
Using the power rule: $\int (y - 3)^{2} d y = \frac{( y - 3 ) ^{3}}{3}$
So, $A = [\frac{( y - 3 ) ^{3}}{3}]_{0}^{3}$

Substitute Upper and Lower Limits

At upper limit $y = 3$ : $\frac{( 3 - 3 ) ^{3}}{3} = 0$
At lower limit $y = 0$ : $\frac{( 0 - 3 ) ^{3}}{3} = \frac{- 27}{3} = - 9$
Area $A = 0 - (- 9) = 9$

Final Answer and Key Takeaway

The area of the bounded region is 9 square units.
Correct Option: (2) (corresponding to value 9).
Tip: Integrating along the $y$ -axis avoided splitting the region into two parts, which would have been necessary if integrating along the $x$ -axis.

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The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The parabola is defined by the equation

(y - 2)^{2} = x - 1

. This is a horizontal parabola opening to the right with its vertex located at

(1, 2)

To find the tangent at the point

(2, 3)

, we differentiate the equation implicitly with respect to

x

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

Substituting the

y

-coordinate of the point of tangency,

y = 3

, we find the slope

m

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

Determining the Tangent Line

Using the point-slope form

y - y_{1} = m (x - x_{1})

with the point

(2, 3)

and slope

m = \frac{1}{2}

, we have:

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

Rearranging this equation to express

x

in terms of

y

for easier integration, we obtain:

x = 2 y - 4

The Master Equation

The region is bounded by the parabola on the right (

x = (y - 2)^{2} + 1

), the tangent line on the left (

x = 2 y - 4

), and the

x

-axis (

y = 0

). The region extends vertically from

y = 0

y = 3

We calculate the area

A

by integrating with respect to

y

A = \int_{0}^{3} (x_{right} - x_{left}) d y

Substituting the functions into the integral:

A = \int_{0}^{3} [((y - 2)^{2} + 1) - (2 y - 4)] d y

Final Calculation

Simplifying the integrand:

(y^{2} - 4 y + 4 + 1) - 2 y + 4 = y^{2} - 6 y + 9 = (y - 3)^{2}

The integral becomes:

A = \int_{0}^{3} (y - 3)^{2} d y

Applying the power rule for integration:

A = [\frac{( y - 3 ) ^{3}}{3}]_{0}^{3}

Evaluating at the limits:

A = (\frac{( 3 - 3 ) ^{3}}{3}) - (\frac{( 0 - 3 ) ^{3}}{3}) = 0 - (\frac{- 27}{3}) = 9

The area of the bounded region is $9$ square units.

Common Traps & FAQs

Why do we integrate with respect to

y

instead of

x

Great question! If you integrate with respect to

x

, you would have to split the region into two parts: one part bounded by the parabola's lower and upper branches, and another part bounded by the tangent line. By integrating with respect to

y

, the boundaries are simply

x = (y - 2)^{2} + 1

on the right and

x = 2 y - 4

on the left, allowing us to compute the area in one clean, elegant step.

How did we get the equation of the tangent line as

x = 2 y - 4

We used the point-slope form of a line,

y - y_{1} = m (x - x_{1})

. We found the slope

m

by differentiating the parabola's equation implicitly, giving us

m = \frac{1}{2}

at the point

(2, 3)

. Substituting these into the formula

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

and rearranging for

x

gives

x = 2 y - 4

Is it possible to get a negative area?

The area itself must be positive. If you get a negative result, it usually means you have swapped the right and left boundaries in your integral setup. Remember, the area is the integral of (right curve - left curve). If you get a negative value, just take its absolute value, but it is better to set it up correctly from the start!

Animated Solution for Mathematics - Definite Integration: The area of the region bounded by the parabola $(y - 2)^{2} = x - 1$ , the tangent of the parabola at the point (2, 3) and the x-axis is:

Select Answer:

Visualized Solution

Visualize the Parabola $(y - 2)^{2} = x - 1$

Differentiate to Find the Slope Function

Calculate Slope at Point $(2, 3)$

Equation of the Tangent Line

Identify the Bounded Region and Limits

Set up the Integral with respect to $y$

Simplify the Algebraic Expression

Integrate the Simplified Function

Substitute Upper and Lower Limits

Final Answer and Key Takeaway

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Determining the Tangent Line

The Master Equation

Final Calculation

Similar Questions

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 :

The area (in sq. units) bounded by the parabola $y = x^{2} - 1$ , the tangent at the point $(2, 3)$ to it and the y-axis is :

The area (in sq. units) of the region bounded by the parabola, $y = x^{2} + 2$ and the lines, $y = x + 1$ , $x = 0$ and $x = 3$ , is :

The area of the region bounded by the curves $y = ∣ x - 2∣, x = 1, x = 3$ and the x-axis is

The area of the region enclosed by the parabola $(y - 2)^{2} = x - 1$ , the line $x - 2 y + 4 = 0$ and the positive coordinate axes is_________.

The area (in square units) of the region bounded by the parabola $y^{2} = 4 (x - 2)$ and the line $y = 2 x - 8$

The area (in sq. units) in the first quadrant bounded by the parabola, $y = x^{2} + 1$ , the tangent to it at the point $(2, 5)$ and the coordinate axes is:

The area of the bounded region enclosed by the curve $y = 3 - x - \frac{1}{2} - ∣ x + 1∣$ and the x-axis is

The area bounded by the curves $y = ∣ x - 1∣ + ∣ x - 2∣$ and $y = 3$ is equal to

The area of the region enclosed by the curve $y = x^{3}$ and its tangent at the point $(- 1, - 1)$ is

Select Answer:

Visualized Solution

Visualize the Parabola .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (y−2)2=x−1

Differentiate to Find the Slope Function

Calculate Slope at Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (2,3)

Equation of the Tangent Line

Identify the Bounded Region and Limits

Set up the Integral with respect to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Simplify the Algebraic Expression

Integrate the Simplified Function

Substitute Upper and Lower Limits

Final Answer and Key Takeaway

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

Determining the Tangent Line

The Master Equation

Final Calculation

Similar Questions

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Visualize the Parabola $(y - 2)^{2} = x - 1$

Calculate Slope at Point $(2, 3)$

Set up the Integral with respect to $y$

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The area (in sq. units) of the region bounded by the parabola, $y = x^{2} + 2$ and the lines, $y = x + 1$ , $x = 0$ and $x = 3$ , is :

The area of the region bounded by the curves $y = ∣ x - 2∣, x = 1, x = 3$ and the x-axis is

The area of the region enclosed by the parabola $(y - 2)^{2} = x - 1$ , the line $x - 2 y + 4 = 0$ and the positive coordinate axes is_________.

The area (in square units) of the region bounded by the parabola $y^{2} = 4 (x - 2)$ and the line $y = 2 x - 8$

The area (in sq. units) in the first quadrant bounded by the parabola, $y = x^{2} + 1$ , the tangent to it at the point $(2, 5)$ and the coordinate axes is:

The area of the bounded region enclosed by the curve $y = 3 - x - \frac{1}{2} - ∣ x + 1∣$ and the x-axis is

The area bounded by the curves $y = ∣ x - 1∣ + ∣ x - 2∣$ and $y = 3$ is equal to

The area of the region enclosed by the curve $y = x^{3}$ and its tangent at the point $(- 1, - 1)$ is