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

Q: Wait, why did we use the section formula on the x-coordinates instead of the y-coordinates?

That is a brilliant observation! You could absolutely use the y-coordinates. The section formula works for both dimensions. Using the x-coordinate is often cleaner because the x-coordinate of point $B$ is $0$, which simplifies the algebra significantly. Try it with the y-coordinates—you will arrive at the exact same differential equation!

Q: Why is the differential equation $xy' + 3y = 0$ and not $xy' - 3y = 0$?

This comes down to the sign convention in the section formula. When we set $x = \frac{3(x_A) + 1(x_B)}{4}$, the term involving $y'$ comes from $x_A$. Because the tangent has a negative slope in the first quadrant to create intercepts on both axes, the algebra naturally forces the sign to be positive. Always trust the algebra when the geometry is set up correctly.

Q: How do I know if I should use $\ln|y|$ or just $\ln y$ during integration?

In JEE problems, we generally assume the domain where the function is defined. Since the curve passes through $(1, 1)$, $y$ is positive. Using the absolute value $\ln|y|$ is the mathematically rigorous way to handle the integral of $1/y$, and it ensures you don't lose the constant of integration $C$.

Q: Is it possible for the tangent to have a positive slope?

If the tangent had a positive slope, it would intersect the x-axis to the left of $P$ and the y-axis below $P$. While the math would still hold, the specific ratio $BP:AP = 3:1$ and the point $(1, 1)$ constrain the curve to a specific region where the slope must be negative. The differential equation captures this behavior perfectly.

Q: Do I always need to check all the options?

In a multiple-correct JEE Advanced question, yes! Even if you find one correct option, the others might also be valid. Always verify your final function $y = 1/x^3$ against every option provided to ensure you haven't missed a hidden truth.

Question

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 $BP:AP = 3:1$, then

Options:

(A) equation of curve is $xy' - 3y = 0$

(B) normal at (1, 1) is $x + 3y = 4$

(C) curve passes through (2, 1/8)

(D) equation of curve is $xy' + 3y = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a smooth, elegant curve defined by $y = f(x)$. At any arbitrary point $P(x, y)$ on this path, you draw a tangent line. This line represents the instantaneous direction of your motion.

As this tangent line slices through the coordinate plane, it carves out two intercepts: $A$ on the x-axis and $B$ on the y-axis. The problem requires us to find the nature of this curve given that the point $P$ divides the segment $BA$ in a strict ratio of $3:1$.

### Defining the Tangent

To begin, we must capture the essence of the tangent line. At any point $P(x, y)$, the slope of the tangent is $y' = \frac{dy}{dx}$.

Using the point-slope form, the equation of our tangent line is $Y - y = y'(X - x)$, where $(X, Y)$ are the coordinates of any point on the line. To find the intercepts, we set one coordinate to zero.

For the x-intercept $A$, we set $Y = 0$, yielding $X = x - \frac{y}{y'}$. Thus, $A = (x - \frac{y}{y'}, 0)$.

Similarly, for the y-intercept $B$, we set $X = 0$, yielding $Y = y - xy'$. Thus, $B = (0, y - xy')$.

### The Power of the Section Formula

Now, we invoke the section formula. We are told that $P(x, y)$ divides $BA$ in the ratio $3:1$. This means $P$ is the weighted average of the endpoints $B$ and $A$.

Focusing on the x-coordinate, we have:

$$x = \frac{3(x_A) + 1(x_B)}{3 + 1}$$

Substituting our expressions for $A$ and $B$, where $x_A = x - \frac{y}{y'}$ and $x_B = 0$, we get:

$$x = \frac{3(x - \frac{y}{y'}) + 0}{4}$$

### The Birth of the Differential Equation

This is where the magic happens. We multiply by $4$ to clear the fraction: $4x = 3x - \frac{3y}{y'}$.

Subtracting $3x$ from both sides leaves us with $x = -\frac{3y}{y'}$. Rearranging this, we arrive at the beautiful, compact differential equation:

$$xy' + 3y = 0$$

This equation is the "DNA" of our curve. It dictates exactly how the slope must change to satisfy the geometric constraint.

### Solving the Mystery

To solve $x \frac{dy}{dx} = -3y$, we use the method of separation of variables. We group all $y$ terms on one side and all $x$ terms on the other:

$$\frac{dy}{y} = -3 \frac{dx}{x}$$

Integrating both sides, we obtain $\ln|y| = -3 \ln|x| + \ln C$. Using the properties of logarithms, this simplifies to $\ln|y| = \ln(C \cdot |x|^{-3})$, which leads us to the general solution:

$$y = \frac{C}{x^3}$$

### Final Calculation

Since the curve passes through $(1, 1)$, we substitute these values to find $C = 1$. Our final curve is **$y = \frac{1}{x^3}$**.

We have successfully navigated from a geometric constraint to a differential equation, and finally to the explicit function. By testing the point $(2, 1/8)$, we see that $1/(2^3) = 1/8$, confirming that the curve indeed satisfies the derived relationship.

Animated Solution for Mathematics - Differential Equations: 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

Select Answer:

Visualized Solution

Visualizing the Geometry

Equation of the Tangent

Finding the x-intercept (Point A)

Finding the y-intercept (Point B)

Applying the Section Formula

Substituting into the Section Formula

Simplifying to a Differential Equation

Solving by Variable Separation

Integration and Logarithmic Properties

Finding the Particular Solution

Verifying the Options

The Sigma Insight: Variable Separable Method

Analyzing the Setup

Defining the Tangent

The Power of the Section Formula

The Birth of the Differential Equation

Solving the Mystery

Final Calculation

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