JEE Main 2022 (24 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: The slope of normal at any point $(x, y), x > 0, y > 0$ on the curve $y = y (x)$ is given by $\frac{x ^{2}}{x y - x ^{2} y ^{2} - 1}$ . If the curve passes through the point $(1, 1)$ , then $e \cdot y (e)$ is equal to

Q: Wait, why is the slope of the normal $-dx/dy$ and not $-dy/dx$?

Great question! The slope of the tangent is $dy/dx$. Since the normal is perpendicular to the tangent, its slope must be the negative reciprocal of the tangent's slope. Therefore, the slope of the normal is $-1 / (dy/dx)$, which is $-dx/dy$. It is a common trap to confuse the two, but always remember the geometric relationship!

Q: Why can we drop the absolute value in $\ln|x|$?

The problem explicitly states that $x > 0$. Since $x$ is always positive, $|x|$ is just $x$. This allows us to simplify $\ln|x|$ to $\ln x$ without any loss of generality.

Select Answer:

Visualized Solution

Visualizing the Curve and Normal

Let the curve be $y = y (x)$ .
Consider any point $P (x, y)$ on the curve.
The normal is perpendicular to the tangent at $P$ .

Slope of the Normal

Slope of Tangent $= \frac{d y}{d x}$
Slope of Normal $= - \frac{1}{Slope of Tangent}$
Slope of Normal $= - \frac{d x}{d y}$

Equating the Slopes

Given Slope of Normal $= \frac{x ^{2}}{x y - x ^{2} y ^{2} - 1}$
Equating the two expressions:
$- \frac{d x}{d y} = \frac{x ^{2}}{x y - x ^{2} y ^{2} - 1}$

Rearranging the Differential Equation

Cross-multiply to remove fractions:
$- (x y - x^{2} y^{2} - 1) d x = x^{2} d y$
Distribute the negative sign:
$(x^{2} y^{2} - x y + 1) d x = x^{2} d y$

Expanding and Grouping Terms

Expand the left side:
$x^{2} y^{2} d x - x y d x + d x = x^{2} d y$
Group terms to isolate $d x$ :
$x^{2} y^{2} d x + d x = x^{2} d y + x y d x$

Identifying the Exact Differential

Factor out $x$ on the right side:
$x^{2} y^{2} d x + d x = x (x d y + y d x)$
Recall the Product Rule of differentiation:
$d (x y) = x d y + y d x$

Substituting the Exact Differential

Substitute $d (x y)$ into the equation:
$x^{2} y^{2} d x + d x = x d (x y)$
Factor out $d x$ on the left side:
$(1 + (x y)^{2}) d x = x d (x y)$

Separating the Variables

Rearrange to separate $x$ and $x y$ :
$\frac{d x}{x} = \frac{d ( x y )}{1 + ( x y ) ^{2}}$
The variables are now completely separated.

Integrating Both Sides

Integrate the separated equation:
$\int \frac{d x}{x} = \int \frac{d ( x y )}{1 + ( x y ) ^{2}}$
Standard integral formulas:
$ln ∣ x ∣ + C = tan^{- 1} (x y)$

Applying Initial Conditions

The curve passes through $(1, 1)$ .
Substitute $x = 1$ and $y = 1$ :
$ln (1) + C = tan^{- 1} (1 \cdot 1)$

Finding the Constant $C$

Evaluate the terms:
$0 + C = tan^{- 1} (1)$
$C = \frac{π}{4}$
Particular Solution:
$ln x + \frac{π}{4} = tan^{- 1} (x y)$

Substituting $x = e$

We need to find the value of $e \cdot y (e)$ .
Substitute $x = e$ into the particular solution:
$ln (e) + \frac{π}{4} = tan^{- 1} (e \cdot y (e))$

Evaluating at $x = e$

Since $ln (e) = 1$ :
$1 + \frac{π}{4} = tan^{- 1} (e \cdot y (e))$
Take tangent on both sides:
$e \cdot y (e) = tan (1 + \frac{π}{4})$

Applying Tangent Addition Formula

Use identity: $tan (A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B}$
$e \cdot y (e) = \frac{t a n ( 1 ) + t a n ( \frac{π}{4} )}{1 - t a n ( 1 ) t a n ( \frac{π}{4} )}$
Since $tan (\frac{π}{4}) = 1$ :
$e \cdot y (e) = \frac{t a n ( 1 ) + 1}{1 - t a n ( 1 ) \cdot 1}$

Final Answer

Rearranging the terms:
$e \cdot y (e) = \frac{1 + t a n ( 1 )}{1 - t a n ( 1 )}$
This matches one of the given options.

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The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

Imagine standing on a smooth, elegant curve on the coordinate plane. At any point

P (x, y)

, we are interested in the normal line, which stands perfectly perpendicular to the tangent.

We are given the slope of this normal as:

\frac{d y}{d x}_{normal} = \frac{x ^{2}}{x y - x ^{2} y ^{2} - 1}

Our goal is to find the curve

y = y (x)

that passes through

(1, 1)

and then evaluate

e \cdot y (e)

Translating Geometry into Algebra

The slope of the tangent is

\frac{d y}{d x}

. Since the normal is perpendicular to the tangent, its slope is the negative reciprocal,

- \frac{d x}{d y}

Equating our theoretical slope to the given expression, we have:

- \frac{d x}{d y} = \frac{x ^{2}}{x y - x ^{2} y ^{2} - 1}

Cross-multiplying gives us:

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

Distributing the negative sign, we obtain the differential equation:

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

The Hidden Symmetry

Expanding and grouping the terms, we have:

x^{2} y^{2} d x - x y d x + d x = x^{2} d y

Moving the

- x y d x

term to the right side yields:

x^{2} y^{2} d x + d x = x^{2} d y + x y d x

Observe the right side:

x^{2} d y + x y d x

. Factoring out an

x

, we get

x (x d y + y d x)

Recognizing that

d (x y) = x d y + y d x

, our equation simplifies to:

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

The Elegance of Separation

We now separate the variables by dividing by

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

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

Integrating both sides results in:

ln x + C = tan^{- 1} (x y)

Given that the curve passes through

(1, 1)

, we substitute

x = 1

and

y = 1

ln 1 + C = tan^{- 1} (1) \Rightarrow 0 + C = \frac{π}{4} \Rightarrow C = \frac{π}{4}

Thus, our particular solution is:

ln x + \frac{π}{4} = tan^{- 1} (x y)

The Final Act

To find

e \cdot y (e)

, we substitute

x = e

into our equation:

ln e + \frac{π}{4} = tan^{- 1} (e \cdot y (e))

Since

ln e = 1

, we have:

1 + \frac{π}{4} = tan^{- 1} (e \cdot y (e))

Taking the tangent of both sides:

e \cdot y (e) = tan (1 + \frac{π}{4})

Using the tangent addition formula

tan (A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B}

, we calculate:

e \cdot y (e) = \frac{tan 1 + tan ( π /4 )}{1 - tan 1 tan ( π /4 )}

Since

tan (π /4) = 1

, the final result is:

e \cdot y (e) = \frac{tan 1 + 1}{1 - tan 1}

Common Traps & FAQs

Wait, why is the slope of the normal

- d x / d y

and not

- d y / d x

Great question! The slope of the tangent is

d y / d x

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

- 1/ (d y / d x)

, which is

- d x / d y

. It is a common trap to confuse the two, but always remember the geometric relationship!

How did you know to group terms to get

d (x y)

This is a classic JEE pattern recognition skill. Whenever you see

x d y

and

y d x

appearing together in a differential equation, it is almost certainly the product rule

d (x y) = x d y + y d x

. Keep an eye out for this; it turns complex equations into simple, integrable forms.

Why can we drop the absolute value in

ln ∣ x ∣

The problem explicitly states that

x > 0

. Since

x

is always positive,

∣ x ∣

is just

x

. This allows us to simplify

ln ∣ x ∣

ln x

without any loss of generality.

Animated Solution for Mathematics - Differential Equations: The slope of normal at any point $(x, y), x > 0, y > 0$ on the curve $y = y (x)$ is given by $\frac{x ^{2}}{x y - x ^{2} y ^{2} - 1}$ . If the curve passes through the point $(1, 1)$ , then $e \cdot y (e)$ is equal to

Select Answer:

Visualized Solution

Visualizing the Curve and Normal

Slope of the Normal

Equating the Slopes

Rearranging the Differential Equation

Expanding and Grouping Terms

Identifying the Exact Differential

Substituting the Exact Differential

Separating the Variables

Integrating Both Sides

Applying Initial Conditions

Finding the Constant $C$

Substituting $x = e$

Evaluating at $x = e$

Applying Tangent Addition Formula

Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Setup

Translating Geometry into Algebra

The Hidden Symmetry

The Elegance of Separation

The Final Act

Similar Questions

Let a smooth curve $y = f (x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\frac{- y}{x}$ . If the curve passes through the point $(1, 2)$ and $(8, 1)$ , then $∣ y (\frac{1}{8}) ∣$ is equal to

A curve $y = f (x)$ passes through (1, 1) and at $P (x, y)$ , tangent cuts the x-axis and y-axis at A and B respectively such that $B P : A P = 3 : 1$ , then

If a curve $y = f (x)$ passes through the point $(1, - 1)$ and satisfies the differential equation, $y (1 + x y) d x = x d y$ , then $f (- \frac{1}{2})$ is equal to :

Given that the slope of the tangent to a curve $y = y (x)$ at any point $(x, y)$ is $2 y / x^{2}$ . If the curve passes through the centre of the circle $x^{2} + y^{2} - 2 x - 2 y = 0$ , then its equation is :

The slope of the tangent to a curve $C : y = y (x)$ at any point $(x, y)$ on it is $\frac{2 e ^{2 x} - 6 e ^{- x} + 9}{2 + 9 e ^{- 2 x}}$ . If $C$ passes through the points $(0, \frac{1}{2} + \frac{π}{2 2})$ and $(α, \frac{1}{2} e^{2 α})$ , then $e^{α}$ is equal to :

Let $Γ$ denote a curve $y = y (x)$ which is in the first quadrant and let the point $(1, 0)$ lie on it. Let the tangent to $Γ$ at a point $P$ intersect the y-axis at $Y_{P}$ . If $P Y_{P}$ has length 1 for each point $P$ on $Γ$ , then which of the following options is/are correct?

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is

Let a curve $y = y (x)$ be given by the solution of the differential equation $cos (\frac{1}{2} cos^{- 1} (e^{- x})) d x = e^{2 x} - 1 d y$ . If it intersects $y$ -axis at $y = - 1$ , and the intersection point of the curve with $x$ -axis is $(α, 0)$ , then $e^{α}$ is equal to

If the curve $y = y (x)$ represented by the solution of the differential equation $(2 x y^{2} - y) d x + x d x = 0$ , passes through the intersection of the lines, $2 x - 3 y = 1$ and $3 x + 2 y = 8$ , then $∣ y (1) ∣$ is equal to

Select Answer:

Visualized Solution

Visualizing the Curve and Normal

Slope of the Normal

Equating the Slopes

Rearranging the Differential Equation

Expanding and Grouping Terms

Identifying the Exact Differential

Substituting the Exact Differential

Separating the Variables

Integrating Both Sides

Applying Initial Conditions

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

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=e

Evaluating at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=e

Applying Tangent Addition Formula

Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Setup

Translating Geometry into Algebra

The Hidden Symmetry

The Elegance of Separation

The Final Act

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a curve y=f(x) passes through the point (1,−1) and satisfies the differential equation, y(1+xy)dx=xdy, then f(−21​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the solution curve y=y(x) of the differential equation (1+y2)(1+loge​x)dx+xdy=0,x>0 passes through the point (1,1) and y(e)=β+tan(3/2)α−tan(3/2)​, then α+2β is

Finding the Constant $C$

Substituting $x = e$

Evaluating at $x = e$

A curve $y = f (x)$ passes through (1, 1) and at $P (x, y)$ , tangent cuts the x-axis and y-axis at A and B respectively such that $B P : A P = 3 : 1$ , then

If a curve $y = f (x)$ passes through the point $(1, - 1)$ and satisfies the differential equation, $y (1 + x y) d x = x d y$ , then $f (- \frac{1}{2})$ is equal to :

If the solution curve $y = y (x)$ of the differential equation $(1 + y^{2}) (1 + lo g_{e} x) d x + x d y = 0, x > 0$ passes through the point $(1, 1)$ and $y (e) = \frac{α - t a n ( 3/2 )}{β + t a n ( 3/2 )}$ , then $α + 2 β$ is