Let y = f(x) be a curve passing through...

MathematicsFormation of Differential EquationsJEE Advanced 1995Moderate

Let $y = f (x)$ be a curve passing through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve lies in the first quadrant and has area 2. Form the differential equation and determine all such possible curves.

Answer:

x + y = 2

and

x y = 1

Visualized Solution (Hindi)

Visualizing the Geometric Constraint

Let the curve be $y = f (x)$ passing through $(1, 1)$ .
At any point $P (x, y)$ , let the slope of the tangent be $y^{'} = \frac{d y}{d x}$ .
The triangle is formed by the coordinate axes and the tangent line.
Given: Area of Triangle $= 2$ and it lies in the First Quadrant.

Equation of the Tangent Line

Equation of tangent at $P (x, y)$ is:
$Y - y = y^{'} (X - x)$
where $y^{'} = \frac{d y}{d x}$ is the slope at point $P$ .

Finding the Intercepts

To find X-intercept, set $Y = 0$ :
$- y = y^{'} (X - x) \Rightarrow X = x - \frac{y}{y ^{'}}$
To find Y-intercept, set $X = 0$ :
$Y - y = y^{'} (- x) \Rightarrow Y = y - x y^{'}$

Setting up the Area Equation

Area of triangle $= \frac{1}{2} ∣ X \cdot Y ∣ = 2$
$\frac{1}{2} ∣ (x - \frac{y}{y ^{'}}) (y - x y^{'}) ∣ = 2$
$∣ (x - \frac{y}{y ^{'}}) (y - x y^{'}) ∣ = 4$

Forming the Differential Equation

Simplifying the product:
$(\frac{x y ^{'} - y}{y ^{'}}) (y - x y^{'}) = 4$
$- \frac{( y - x y ^{'} ) ^{2}}{y ^{'}} = 4$
The Differential Equation is: $(y - x y^{'})^{2} = - 4 y^{'}$

Solving for General Solution

Let $y - x y^{'} = c$ . Then $c^{2} = - 4 y^{'} \Rightarrow y^{'} = - \frac{c ^{2}}{4}$ .
Substitute $y^{'}$ back: $y - x (- \frac{c ^{2}}{4}) = c$
General Solution: $y + \frac{x c ^{2}}{4} = c$

Finding the First Curve: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x + y = 2

Passes through $(1, 1)$ : $1 + \frac{1 \cdot c ^{2}}{4} = c$
$c^{2} - 4 c + 4 = 0 \Rightarrow (c - 2)^{2} = 0 \Rightarrow c = 2$
Substituting $c = 2$ into $y + \frac{x c ^{2}}{4} = c$ :
$y + x = 2 \Rightarrow x + y = 2$

Finding the Singular Solution

For singular solution, differentiate $y = c - \frac{x c ^{2}}{4}$ w.r.t. $c$ :
$0 = 1 - \frac{2 x c}{4} \Rightarrow c = \frac{2}{x}$
Substitute $c = \frac{2}{x}$ back into the general solution:
$y = \frac{2}{x} - \frac{x}{4} (\frac{2}{x})^{2} = \frac{2}{x} - \frac{1}{x} = \frac{1}{x}$
This gives the curve $x y = 1$ .

Conclusion and Final Curves

The two possible curves are:
1. The straight line: $x + y = 2$
2. The rectangular hyperbola: $x y = 1$
Key Takeaway: Always check for singular solutions in non-linear differential equations of Clairaut's form.

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Visualized Solution (Hindi)

Visualizing the Geometric Constraint

Equation of the Tangent Line

Finding the Intercepts

Setting up the Area Equation

Forming the Differential Equation

Solving for General Solution

Finding the First Curve: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x+y=2

Finding the Singular Solution

Conclusion and Final Curves

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