JEE Main 2021 (27 Aug Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Definite Integration: 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 :

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 because the lower boundary changes at the vertex. By integrating with respect to $y$, the right boundary is always the parabola and the left boundary is always the tangent line, making it a single, elegant integral.

Q: How did you get the tangent equation $x = 2y - 4$?

We first found the point of tangency by substituting $y = 3$ into the parabola equation, giving $x = 2$. Then, we found the derivative $\frac{dx}{dy} = 2(y - 2)$ (or $\frac{dy}{dx} = \frac{1}{2(y-2)}$). At $y = 3$, the slope $\frac{dx}{dy} = 2(3 - 2) = 2$. Using the point-slope form $x - x_1 = \frac{dx}{dy}(y - y_1)$, we get $x - 2 = 2(y - 3)$, which simplifies to $x = 2y - 4$.

Select Answer:

Visualized Solution

Visualizing the Parabola

Given Parabola: $(y - 2)^{2} = (x - 1)$
Vertex at $(1, 2)$ , opens rightwards.
Objective: Find the area bounded by this parabola, its tangent at $y = 3$ , and the $x$ -axis.

Finding the Point of Tangency

We need the tangent at the point where the ordinate is $3$ .
Substitute $y = 3$ into the parabola's equation:
$(3 - 2)^{2} = x - 1$
$1 = x - 1 \Rightarrow x = 2$
Point of Tangency: $P (2, 3)$

Calculating the Slope of Tangent

To find the tangent's equation, we first need its slope, $m = \frac{d y}{d x}$ .
Differentiate $(y - 2)^{2} = x - 1$ with respect to $x$ :
$2 (y - 2) \cdot \frac{d y}{d x} = 1$
$\frac{d y}{d x} = \frac{1}{2 ( y - 2 )}$

Slope at $P (2, 3)$

Substitute the coordinates of $P (2, 3)$ into the derivative.
$m = \frac{d y}{d x}_{(2, 3)} = \frac{1}{2 ( 3 - 2 )}$
$m = \frac{1}{2 ( 1 )} = \frac{1}{2}$

Equation of the Tangent Line

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

Identifying the Bounded Region

The region is bounded by:
1. Parabola: $x = (y - 2)^{2} + 1$ (Right boundary)
2. Tangent: $x = 2 y - 4$ (Left boundary)
3. $x$ -axis: $y = 0$ (Lower boundary)
Upper boundary is the point of tangency at $y = 3$ .

Setting up the Area Integral

Integrating with respect to $y$ (horizontal strips) is much easier here.
Area $A = \int_{y_{min}}^{y_{ma x}} (x_{R i g h t} - x_{L e f t}) d y$
$x_{R} = (y - 2)^{2} + 1$
$x_{L} = 2 y - 4$
Limits: $y = 0$ to $y = 3$

Substituting the Boundaries

$A = \int_{0}^{3} [((y - 2)^{2} + 1) - (2 y - 4)] d y$
Let's expand the terms inside the integral carefully.

Simplifying the Integrand

Expand $(y - 2)^{2}$ : $y^{2} - 4 y + 4$
Substitute back: $A = \int_{0}^{3} (y^{2} - 4 y + 4 + 1 - 2 y + 4) d y$
Combine like terms: $A = \int_{0}^{3} (y^{2} - 6 y + 9) d y$
Notice a perfect square: $y^{2} - 6 y + 9 = (y - 3)^{2}$

Performing the Integration

$A = \int_{0}^{3} (y - 3)^{2} d y$
Use the power rule for integration: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1}$
$A = [\frac{( y - 3 ) ^{3}}{3}]_{0}^{3}$

Evaluating the Limits

Upper limit ( $y = 3$ ): $\frac{( 3 - 3 ) ^{3}}{3} = 0$
Lower limit ( $y = 0$ ): $\frac{( 0 - 3 ) ^{3}}{3} = \frac{- 27}{3} = - 9$
$A = Upper - Lower$
$A = 0 - (- 9) = 9$

Final Answer & Takeaway

The area of the bounded region is 9 sq. units.
Pro Tip: Always check if integrating with respect to $y$ (horizontal strips) avoids splitting the integral into multiple parts.

00:00 / 00:00

The Sigma Insight: Area Bounded by Curves

Solution Diagram

Analyzing the Geometry

The parabola is defined by the equation

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

. Because the squared term involves

y

, this parabola opens horizontally to the right with its vertex located at

(1, 2)

We are tasked with finding the area of the region bounded by this curve, a specific tangent line, and the

x

-axis.

The Tangent's Dance

First, we identify the point of tangency where the ordinate is

y = 3

. Substituting this into the parabola equation:

(3 - 2)^{2} = x - 1 ⟹ 1 = x - 1 ⟹ x = 2

Thus, our point of tangency is

P (2, 3)

. To find the slope

m

, we differentiate the equation

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

with respect to

x

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

At point

P (2, 3)

, the slope is

m = \frac{1}{2 ( 3 - 2 )} = \frac{1}{2}

. Using the point-slope form

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

, we obtain:

y - 3 = \frac{1}{2} (x - 2) ⟹ 2 y - 6 = x - 2 ⟹ x = 2 y - 4

The Strategic Choice

Integrating with respect to

x

would require splitting the region into two parts. However, integrating with respect to

y

allows for a single, continuous integral.

The right boundary is the parabola

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

, and the left boundary is the tangent line

x = 2 y - 4

. The region spans from the

x

-axis (

y = 0

) to the point of tangency (

y = 3

The area

A

is given by the integral:

A = \int_{0}^{3} (x_{R i g h t} - x_{L e f t}) d y

The Integration

Substituting the expressions for

x

into the integral:

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

Expanding the integrand:

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

Now, we evaluate the simplified integral:

A = \int_{0}^{3} (y - 3)^{2} d y = [\frac{( y - 3 ) ^{3}}{3}]_{0}^{3}

Evaluating at the boundaries:

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

The area of the 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 because the lower boundary changes at the vertex. By integrating with respect to

y

, the right boundary is always the parabola and the left boundary is always the tangent line, making it a single, elegant integral.

How did you get the tangent equation

x = 2 y - 4

We first found the point of tangency by substituting

y = 3

into the parabola equation, giving

x = 2

. Then, we found the derivative

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

(or

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

). At

y = 3

, the slope

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

. Using the point-slope form

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

, we get

x - 2 = 2 (y - 3)

, which simplifies to

x = 2 y - 4

Why is the lower limit of integration

0

The problem asks for the area bounded by the parabola, the tangent, and the

x

-axis. The

x

-axis is defined by the equation

y = 0

. Since the region is bounded by this line,

y = 0

is our lower limit of integration.

Animated Solution for Mathematics - Definite Integration: 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 :

Select Answer:

Visualized Solution

Visualizing the Parabola

Finding the Point of Tangency

Calculating the Slope of Tangent

Slope at $P (2, 3)$

Equation of the Tangent Line

Identifying the Bounded Region

Setting up the Area Integral

Substituting the Boundaries

Simplifying the Integrand

Performing the Integration

Evaluating the Limits

Final Answer & Takeaway

The Sigma Insight: Area Bounded by Curves

Analyzing the Geometry

The Tangent's Dance

The Strategic Choice

The Integration

Similar Questions

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:

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

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

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

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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

A line passing through the point $A (- 2, 0)$ , touches the parabola $P : y^{2} = x - 2$ at the point $B$ in the first quadrant. The area, of the region bounded by the line $A B$ , parabola $P$ and the $x$ -axis, is :-

Select Answer:

Visualized Solution

Visualizing the Parabola

Finding the Point of Tangency

Calculating the Slope of Tangent

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

Equation of the Tangent Line

Identifying the Bounded Region

Setting up the Area Integral

Substituting the Boundaries

Simplifying the Integrand

Performing the Integration

Evaluating the Limits

Final Answer & Takeaway

The Sigma Insight: Area Bounded by Curves

Analyzing the Geometry

The Tangent's Dance

The Strategic Choice

The Integration

Similar Questions

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Slope at $P (2, 3)$

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:

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