JEE Advanced 1999

LEVELJEE Advanced

Animated Solution for Mathematics - Differential Equations: A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve.

Q: Why did we use the substitution $y = vx$ instead of trying to separate variables immediately?

Because the differential equation $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$ is not separable in its current form. Notice that the numerator and denominator are both homogeneous functions of degree 2. This is the hallmark of a homogeneous differential equation. The substitution $y = vx$ is the 'magic key' that transforms this complex expression into a separable one, allowing us to isolate $v$ and $x$.

Progressive Hints

Hint 1

Visualize the normal line at a generic point

P (x, y)

on the curve. The slope of the tangent is

y^{'} = \frac{d y}{d x}

, so the slope of the normal is

- \frac{1}{y ^{'}}

. Using the point-slope form, construct the equation of the normal line. Recall that the perpendicular distance from the origin

(0, 0)

to a line

A x + B y + C = 0

is given by

\frac{∣ C ∣}{A ^{2} + B ^{2}}

. Can you set up this distance expression?

Hint 2

Once you have the expression for the distance from the origin, equate it to the distance of point

P

from the x-axis, which is simply

∣ y ∣

. You will have an equation involving

x, y,

and

y^{'}

. To simplify this, square both sides to eliminate the square root and the absolute value signs. What does the resulting differential equation look like after you expand and cancel terms?

Hint 3

After simplifying, you should arrive at the differential equation

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

. This is a classic homogeneous differential equation. Use the substitution

y = v x

, which implies

\frac{d y}{d x} = v + x \frac{d v}{d x}

. Substitute this into your equation, simplify the right-hand side, and separate the variables to integrate.

Visualized Solution

Visualizing the Geometric Setup

Let the point on the curve be $P (x, y)$ .
The curve passes through $(1, 1)$ .

Equation of the Normal

Slope of tangent at $P$ is $y^{'} = \frac{d y}{d x}$ .
Slope of normal is $- \frac{1}{y ^{'}}$ .
Equation of normal: $X + y^{'} Y - (x + y y^{'}) = 0$ .

Distance from Origin

Perpendicular distance from $(0, 0)$ to the normal is $d_{1} = \frac{∣ x + y y ^{'} ∣}{1 + ( y ^{'} ) ^{2}}$ .

Distance from x-axis

Distance of $P (x, y)$ from the x-axis is $d_{2} = ∣ y ∣$ .

Equating the Distances

Given condition: $d_{1} = d_{2}$ .
Therefore, $\frac{∣ x + y y ^{'} ∣}{1 + ( y ^{'} ) ^{2}} = ∣ y ∣$ .

Squaring and Expanding

Square both sides: $\frac{( x + y y ^{'} ) ^{2}}{1 + ( y ^{'} ) ^{2}} = y^{2}$ .
Cross-multiply and expand: $x^{2} + 2 x y y^{'} + y^{2} (y^{'})^{2} = y^{2} + y^{2} (y^{'})^{2}$ .

Forming the Differential Equation

Cancel $y^{2} (y^{'})^{2}$ to get $x^{2} + 2 x y y^{'} = y^{2}$ .
Rearrange to isolate $y^{'}$ : $\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{2 x y}$ .
This is a homogeneous differential equation.

Substitution for Homogeneous DE

Substitute $y = v x$ .
Differentiating with respect to $x$ gives $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .

Substituting into the DE

Substitute into DE: $v + x \frac{d v}{d x} = \frac{v ^{2} x ^{2} - x ^{2}}{2 x ( v x )} = \frac{v ^{2} - 1}{2 v}$ .
Rearranging gives $x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v} - v = - \frac{v ^{2} + 1}{2 v}$ .

Separating Variables

Rearrange to separate $v$ and $x$ : $\frac{2 v}{v ^{2} + 1} d v = - \frac{1}{x} d x$ .

Integrating Both Sides

Integrate: $\int \frac{2 v}{v ^{2} + 1} d v = - \int \frac{1}{x} d x$ .
This yields $ln (v^{2} + 1) = - ln ∣ x ∣ + ln C$ .
Simplifies to $v^{2} + 1 = \frac{C}{x}$ .

Reverting to $x$ and $y$

Substitute $v = \frac{y}{x}$ : $\frac{y ^{2}}{x ^{2}} + 1 = \frac{C}{x}$ .
Multiply by $x^{2}$ : $x^{2} + y^{2} = C x$ .

Finding the Specific Curve

The curve passes through $(1, 1)$ .
Substitute $x = 1, y = 1$ : $1^{2} + 1^{2} = C (1) \Rightarrow C = 2$ .
Final equation: $x^{2} + y^{2} - 2 x = 0$ .
(Also $x = 1$ is a valid solution).

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The Sigma Insight: Homogeneous Differential Equations

Solution Diagram

Analyzing the Setup

We are tasked with finding the equation of a curve passing through the point

(1, 1)

. The geometric condition provided is that the perpendicular distance from the origin to the normal at any point

P (x, y)

equals the distance of

P

from the x-axis.

Let the slope of the tangent at

P

y^{'} = \frac{d y}{d x}

. Consequently, the slope of the normal line, which is perpendicular to the tangent, is

- \frac{1}{y ^{'}}

Constructing the Normal

Using the point-slope form of a line, the equation of the normal passing through

(x, y)

with slope

- \frac{1}{y ^{'}}

is:

Y - y = - \frac{1}{y ^{'}} (X - x)

Rearranging this into the standard form

A X + B Y + C = 0

by multiplying by

y^{'}

, we obtain:

X + y^{'} Y - (x + y y^{'}) = 0

This equation represents the normal line at any arbitrary point

P (x, y)

on the curve.

The Distance Condition

The perpendicular distance

d_{1}

from the origin

(0, 0)

to the normal line is given by the formula

d = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}}

. Substituting our values, we get:

d_{1} = \frac{∣0 + 0 - ( x + y y ^{'} ) ∣}{1 ^{2} + ( y ^{'} ) ^{2}} = \frac{∣ x + y y ^{'} ∣}{1 + ( y ^{'} ) ^{2}}

The distance

d_{2}

of point

P (x, y)

from the x-axis is simply

∣ y ∣

. Setting

d_{1} = d_{2}

, we have:

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

The Algebraic Dance

To simplify, we square both sides to eliminate the square root and absolute values:

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

Cross-multiplying yields:

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

Expanding both sides results in

x^{2} + 2 x y y^{'} + y^{2} (y^{'})^{2} = y^{2} + y^{2} (y^{'})^{2}

. Canceling the

y^{2} (y^{'})^{2}

term from both sides, we are left with:

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

Isolating the derivative

\frac{d y}{d x}

, we obtain the differential equation:

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

The Homogeneous Transformation

This is a homogeneous differential equation. We apply the substitution

y = v x

, which implies

\frac{d y}{d x} = v + x \frac{d v}{d x}

. Substituting these into the equation:

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

Subtracting

v

from both sides gives:

x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v} - v = \frac{v ^{2} - 1 - 2 v ^{2}}{2 v} = - \frac{v ^{2} + 1}{2 v}

Integration and the Final Reveal

Separating the variables, we have:

\frac{2 v}{v ^{2} + 1} d v = - \frac{1}{x} d x

Integrating both sides yields:

ln (v^{2} + 1) = - ln ∣ x ∣ + ln C

This simplifies to

v^{2} + 1 = \frac{C}{x}

. Substituting

v = \frac{y}{x}

back into the expression:

\frac{y ^{2}}{x ^{2}} + 1 = \frac{C}{x} \Rightarrow y^{2} + x^{2} = C x

Given the curve passes through

(1, 1)

, we substitute

x = 1

and

y = 1

to find

1^{2} + 1^{2} = C (1)

, which means

C = 2

. The final equation of the curve is:

$x^{2} + y^{2} - 2 x = 0$

Common Traps & FAQs

Wait, when we square both sides of the equation

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

, are we introducing extraneous solutions? How do we know if we lost information?

That is a sharp observation! Squaring is a standard technique to remove radicals and absolute values, but you are right to be cautious. While it can introduce extraneous roots, in this context, we are looking for a curve that satisfies the geometric condition. Since the original condition involves absolute values, squaring effectively covers both positive and negative cases. Always check your final result against the original condition to ensure it holds true. In this problem, the squaring process is mathematically sound for finding the family of curves.

Why did we use the substitution

y = v x

instead of trying to separate variables immediately?

Because the differential equation

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

is not separable in its current form. Notice that the numerator and denominator are both homogeneous functions of degree 2. This is the hallmark of a homogeneous differential equation. The substitution

y = v x

is the 'magic key' that transforms this complex expression into a separable one, allowing us to isolate

v

and

x

I got the circle

x^{2} + y^{2} - 2 x = 0

, but the answer key mentions

x = 1

as well. Where did that come from?

That is a brilliant catch! When we performed the division or rearrangement during the solution, we implicitly assumed

y \neq = 0

or specific conditions on the slope. The vertical line

x = 1

is a singular solution. If you check the geometry, the normal to the vertical line

x = 1

is a horizontal line

y = k

. The distance from the origin to this normal is

∣ k ∣

, and the distance of the point

(1, k)

from the x-axis is also

∣ k ∣

. Thus,

x = 1

satisfies the condition perfectly!

Animated Solution for Mathematics - Differential Equations: A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve.

Visualized Solution

Visualizing the Geometric Setup

Equation of the Normal

Distance from Origin

Distance from x-axis

Equating the Distances

Squaring and Expanding

Forming the Differential Equation

Substitution for Homogeneous DE

Substituting into the DE

Separating Variables

Integrating Both Sides

Reverting to $x$ and $y$

Finding the Specific Curve

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

Constructing the Normal

The Distance Condition

The Algebraic Dance

The Homogeneous Transformation

Integration and the Final Reveal

Similar Questions

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Animated Solution for Mathematics - Differential Equations: A curve passing through the point (1, 1) has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x-axis. Determine the equation of the curve.

Visualized Solution

Visualizing the Geometric Setup

Equation of the Normal

Distance from Origin

Distance from x-axis

Equating the Distances

Squaring and Expanding

Forming the Differential Equation

Substitution for Homogeneous DE

Substituting into the DE

Separating Variables

Integrating Both Sides

Reverting to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Finding the Specific Curve

The Sigma Insight: Homogeneous Differential Equations

Analyzing the Setup

Constructing the Normal

The Distance Condition

The Algebraic Dance

The Homogeneous Transformation

Integration and the Final Reveal

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