Let y = f(x) be a curve passing through (1, 1) such that the triangle formed by the coordinate axes

Q: Wait, why do we need the absolute value sign in the area formula? Can't we just ignore it?

That is a fantastic question! In the first quadrant, the intercepts are positive, but the slope $y'$ is negative for a decreasing curve. Without the absolute value, the product of intercepts might come out negative, leading to a sign error in your differential equation. In geometry, area is always positive, so we must respect the physical reality of the triangle by using the absolute value. Always let the geometry guide your algebra!

Q: Why do we have to look for a 'singular solution' after finding the general solution? Isn't the general solution enough?

In non-linear differential equations, the general solution represents a family of curves, but it doesn't always capture the 'envelope'—the curve that touches every member of that family. The singular solution is that hidden envelope. If you stop at the general solution, you are missing half the story! In JEE Advanced, missing the singular solution is a classic trap. Always perform the partial derivative test with respect to the constant $c$ to see if a singular solution exists.

Q: Is this equation actually Clairaut's form? It looks a bit different.

You have a sharp eye! It is a variation of Clairaut's form. Clairaut's equation is typically $y = xy' + f(y')$. Our equation $(y - xy')^2 = -4y'$ can be rearranged to $y = xy' \pm 2\sqrt{-y'}$. It belongs to the same family of equations where the general solution is found by replacing $y'$ with a constant $c$, and the singular solution is found by differentiating with respect to that constant. It is a beautiful piece of mathematical symmetry!

Question

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.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To begin, we need the mathematical language of this tangent. Let the slope of our curve at point $P(x, y)$ be $y' = \frac{dy}{dx}$. Using the point-slope form of a line, the equation of our tangent is:

$$Y - y = y'(X - x)$$

Here, $X$ and $Y$ are the coordinates of any point on the tangent line. To find the triangle's base and height, we need the intercepts.

For the X-intercept, we set $Y = 0$:

$$-y = y'(X - x) \Rightarrow X = x - \frac{y}{y'}$$

For the Y-intercept, we set $X = 0$:

$$Y - y = y'(-x) \Rightarrow Y = y - xy'$$

These two values, $X = x - \frac{y}{y'}$ and $Y = y - xy'$, are the legs of our right-angled triangle. The area is given by $\frac{1}{2} |X \cdot Y| = 2$. Substituting our intercepts, we get:

$$\frac{1}{2} \left| \left( x - \frac{y}{y'} \right) (y - xy') \right| = 2$$

### The Differential Equation

Let us simplify this. Notice that $x - \frac{y}{y'} = \frac{xy' - y}{y'} = -\frac{y - xy'}{y'}$. Substituting this back into our area equation:

$$\frac{1}{2} \left| \left( -\frac{y - xy'}{y'} \right) (y - xy') \right| = 2$$

$$\left| -\frac{(y - xy')^2}{y'} \right| = 4$$

Since we are in the first quadrant and the area is positive, we can drop the absolute value and rearrange to find our differential equation:

$$(y - xy')^2 = -4y'$$

This is the heart of the problem. It is a non-linear differential equation, and it looks intimidating. We have a powerful tool for equations of this form: the substitution method.

### The General Solution

Let us define a constant $c$ such that $y - xy' = c$. Squaring both sides, we get $c^2 = -4y'$, which implies $y' = -\frac{c^2}{4}$. Now, substitute this back into our definition:

$$y - x\left( -\frac{c^2}{4} \right) = c \Rightarrow y = c - \frac{xc^2}{4}$$

This is the general solution, representing a family of lines. We are given that the curve passes through $(1, 1)$. Let us plug these coordinates in:

$$1 = c - \frac{1 \cdot c^2}{4} \Rightarrow 4 = 4c - c^2 \Rightarrow c^2 - 4c + 4 = 0$$

This factors beautifully into $(c - 2)^2 = 0$, giving us $c = 2$. Substituting $c = 2$ back into our general solution, we find our first curve:

$$y = 2 - \frac{x(2)^2}{4} \Rightarrow y = 2 - x \Rightarrow x + y = 2$$

This is a straight line, and it perfectly satisfies the condition. In the world of non-linear differential equations, we must always check for the singular solution—the envelope.

### The Singular Solution

To find the singular solution, we treat the general solution $y = c - \frac{xc^2}{4}$ as a function of $c$ and differentiate with respect to $c$, setting the result to zero:

$$\frac{dy}{dc} = 0 = 1 - \frac{2xc}{4} \Rightarrow 1 - \frac{xc}{2} = 0 \Rightarrow c = \frac{2}{x}$$

Now, we substitute this value of $c$ back into our general solution to find the envelope:

$$y = \left( \frac{2}{x} \right) - \frac{x}{4} \left( \frac{2}{x} \right)^2$$

$$y = \frac{2}{x} - \frac{x}{4} \cdot \frac{4}{x^2} = \frac{2}{x} - \frac{1}{x} = \frac{1}{x}$$

Thus, **$xy = 1$**. This is our second curve, a rectangular hyperbola.

### Final Conclusion

We have found two distinct curves: the straight line **$x + y = 2$** and the hyperbola **$xy = 1$**. Both pass through $(1, 1)$, and both satisfy the geometric constraint of the tangent triangle area. This problem teaches us a vital lesson: never settle for the first answer you find. Always look deeper, check for singular solutions, and trust the geometry.

Visualized Solution

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

Finding the Singular Solution

The Second Curve

Conclusion

The Sigma Insight: Formation of Differential Equations

Analyzing the Setup

The Differential Equation

The General Solution

The Singular Solution

Final Conclusion

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