JEE Advanced 1996

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: A curve $y = f (x)$ passes through the point $P (1, 1)$ . The normal to the curve at $P$ is $a (y - 1) + (x - 1) = 0$ . If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, determine the equation of the curve. Also obtain the area bounded by the $y$ -axis, the curve and the normal to the curve at $P$ .

Q: Wait, why is the slope of the tangent $a$ and not $-1/a$?

Great question! The equation of the normal is $a(y - 1) + (x - 1) = 0$. Rearranging this gives $y - 1 = -\frac{1}{a}(x - 1)$, which means the slope of the normal is $-\frac{1}{a}$. Since the tangent is perpendicular to the normal, its slope must be the negative reciprocal of the normal's slope. Thus, $m_t = -1 / (-1/a) = a$. Keep that perpendicularity rule in your toolkit!

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the geometry at point

P (1, 1)

. The normal line

a (y - 1) + (x - 1) = 0

is the key. Can you rearrange this into the slope-intercept form

y = m x + c

to extract the slope of the normal? Remember, the normal is perpendicular to the tangent.

Hint 2

Hint 2: The Arsenal. Once you have the slope of the normal (

m_{n}

), use the fundamental property of perpendicular lines:

m_{t} \cdot m_{n} = - 1

to find the slope of the tangent (

m_{t}

). Then, translate the phrase 'slope of the tangent is proportional to the ordinate' into the differential equation

\frac{d y}{d x} = k y

Hint 3

Hint 3: The Execution. With the differential equation

\frac{d y}{d x} = a y

established, separate the variables to get

\frac{d y}{y} = a d x

. Integrate both sides to find the general solution, and use the initial condition

f (1) = 1

to determine the constant of integration

C

Visualized Solution

Identify the Point $P (1, 1)$

Given a curve $y = f (x)$ passing through point $P (1, 1)$ .
This means $f (1) = 1$ , which will be our initial condition.

Analyze the Normal at $P$

The equation of the normal at $P$ is given as: $a (y - 1) + (x - 1) = 0$ .
Rearranging into slope-intercept form: $y - 1 = - \frac{1}{a} (x - 1)$ .
Therefore, the slope of the normal is $m_{n} = - \frac{1}{a}$ .

Find the Tangent Slope at $P$

The tangent is perpendicular to the normal, so $m_{t} \times m_{n} = - 1$ .
$m_{t} = - \frac{1}{m _{n}} = - \frac{1}{- 1/ a} = a$ .
Thus, at point $P (1, 1)$ , the derivative $\frac{d y}{d x} = a$ .

Set Up the Differential Equation

The problem states: slope of the tangent at any point is proportional to the ordinate.
Mathematically: $\frac{d y}{d x} \propto y ⟹ \frac{d y}{d x} = k y$ .
We need to find the constant of proportionality, $k$ .

Evaluate the Constant $k$

We know that at $P (1, 1)$ , $y = 1$ and $\frac{d y}{d x} = a$ .
Substituting these into our equation: $a = k (1)$ .
Therefore, $k = a$ .
The exact differential equation is: $\frac{d y}{d x} = a y$ .

Solve by Variable Separation

We have a first-order separable differential equation: $\frac{d y}{d x} = a y$ .
Separate the variables $x$ and $y$ : $\frac{d y}{y} = a d x$ .
Integrate both sides: $\int \frac{d y}{y} = \int a d x$ .

Integrate and Find $C$

Integrating yields: $ln ∣ y ∣ = a x + C$ .
Use the initial condition $P (1, 1)$ to find $C$ : $ln (1) = a (1) + C$ .
Since $ln (1) = 0$ , we get $0 = a + C ⟹ C = - a$ .

Final Equation of the Curve

Substitute $C = - a$ back into the equation: $ln ∣ y ∣ = a x - a = a (x - 1)$ .
Taking the exponential of both sides: $y = e^{a (x - 1)}$ .
This is the required equation of the curve.

Setup the Area Integral

We need the area bounded by the $y$ -axis ( $x = 0$ ), the curve $y = e^{a (x - 1)}$ , and the normal $y = 1 - \frac{1}{a} (x - 1)$ .
The area $A$ is given by the integral of (Upper Curve - Lower Curve) from $x = 0$ to $x = 1$ .
$A = \int_{0}^{1} [(1 - \frac{1}{a} (x - 1)) - e^{a (x - 1)}] d x$ .

Evaluate the Definite Integral

Find the antiderivative: $[x - \frac{1}{2 a} (x - 1)^{2} - \frac{1}{a} e^{a (x - 1)}]_{0}^{1}$ .
Substitute upper limit ( $x = 1$ ): $1 - 0 - \frac{1}{a} = 1 - \frac{1}{a}$ .
Substitute lower limit ( $x = 0$ ): $0 - \frac{1}{2 a} (- 1)^{2} - \frac{1}{a} e^{- a} = - \frac{1}{2 a} - \frac{e ^{- a}}{a}$ .

Final Area Expression

Subtract the lower limit value from the upper limit value.
$A = (1 - \frac{1}{a}) - (- \frac{1}{2 a} - \frac{e ^{- a}}{a})$ .
Simplifying: $A = 1 - \frac{1}{2 a} + \frac{e ^{- a}}{a}$ square units.

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The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

We begin with a curve

y = f (x)

that passes through the anchor point

P (1, 1)

. The normal to the curve at this point is given by the equation

a (y - 1) + (x - 1) = 0

By rearranging this equation into the slope-intercept form, we obtain:

y - 1 = - \frac{1}{a} (x - 1)

This reveals that the slope of the normal at point

P

m_{n} = - \frac{1}{a}

The Tangent-Normal Connection

We invoke the fundamental rule of perpendicularity: the product of the slopes of the tangent and the normal is

- 1

. Given

m_{t} \cdot m_{n} = - 1

, we calculate the slope of the tangent at

P

m_{t} = a

This realization is profound. It confirms that at the point

(1, 1)

, the derivative of the curve is exactly:

\frac{d y}{d x}_{(1, 1)} = a

The Heartbeat of the Curve

The problem states that the slope of the tangent at any point is proportional to the ordinate

y

. Mathematically, this is expressed as:

\frac{d y}{d x} = k y

Substituting our known values at

x = 1

and

y = 1

, we find

a = k (1)

, which implies

k = a

. Thus, our differential equation is defined as:

\frac{d y}{d x} = a y

The Integration Bridge

To solve the differential equation, we use the method of separation of variables. We group the

y

terms and

x

terms:

\frac{d y}{y} = a d x

Integrating both sides yields

ln ∣ y ∣ = a x + C

. We use the anchor point

P (1, 1)

to determine the constant

C

ln (1) = a (1) + C \Rightarrow 0 = a + C \Rightarrow C = - a

Substituting

C

back into the equation, we get

ln ∣ y ∣ = a (x - 1)

. Exponentiating both sides, we arrive at the solution:

y = e^{a (x - 1)}

The Final Harvest

Area

We seek the area bounded by the

y

-axis (

x = 0

), the curve, and the normal. The area is the integral of the upper function minus the lower function from

x = 0

x = 1

The integral is defined as:

A = \int_{0}^{1} [(1 - \frac{1}{a} (x - 1)) - e^{a (x - 1)}] d x

Evaluating this term by term, we compute:

[x - \frac{1}{2 a} (x - 1)^{2} - \frac{1}{a} e^{a (x - 1)}]_{0}^{1}

Substituting the limits of integration, we find the final area to be:

$A = 1 - \frac{1}{2 a} + \frac{e ^{- a}}{a}$

Common Traps & FAQs

Wait, why is the slope of the tangent

a

and not

- 1/ a

Great question! The equation of the normal is

a (y - 1) + (x - 1) = 0

. Rearranging this gives

y - 1 = - \frac{1}{a} (x - 1)

, which means the slope of the normal is

- \frac{1}{a}

. Since the tangent is perpendicular to the normal, its slope must be the negative reciprocal of the normal's slope. Thus,

m_{t} = - 1/ (- 1/ a) = a

. Keep that perpendicularity rule in your toolkit!

Why can't we just use energy conservation to solve this?

I love the enthusiasm for physics, but this is a pure calculus and geometry problem! Energy conservation applies to physical systems with forces and motion. Here, we are dealing with the geometric properties of a curve defined by its slope. We must rely on differential equations to find the function

f (x)

How do I know which function is the 'upper' curve for the area integral?

Excellent intuition. To determine the upper curve, you can test a point in the interval

(0, 1)

, like

x = 0.5

. Calculate the

y

-value for both the normal line and the exponential curve at that point. The one with the higher

y

-value is your upper boundary. In this case, the normal line sits above the curve in the region of interest.

Visualized Solution

Identify the Point $P (1, 1)$

Analyze the Normal at $P$

Find the Tangent Slope at $P$

Set Up the Differential Equation

Evaluate the Constant $k$

Solve by Variable Separation

Integrate and Find $C$

Final Equation of the Curve

Setup the Area Integral

Evaluate the Definite Integral

Final Area Expression

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Tangent-Normal Connection

The Heartbeat of the Curve

The Integration Bridge

The Final Harvest

Similar Questions

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Visualized Solution

Identify the Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(1,1)

Analyze the Normal at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Find the Tangent Slope at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Set Up the Differential Equation

Evaluate the Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Solve by Variable Separation

Integrate and Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Final Equation of the Curve

Setup the Area Integral

Evaluate the Definite Integral

Final Area Expression

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Tangent-Normal Connection

The Heartbeat of the Curve

The Integration Bridge

The Final Harvest

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