Let a curve y = f(x), x ∈ (0, ∞) pass through the points P(1, 3/2) and Q(a, 1/2). If the tangent at

Q: Wait, why do we replace $b$ with $x$ and $f(b)$ with $y$ in the differential equation?

That is a fantastic question! The point $R(b, f(b))$ was just a representative point on the curve. By replacing $b$ with $x$ and $f(b)$ with $y$, we are essentially saying, 'This property must hold true for *every* point on the curve.' This is how we transition from a specific geometric condition at one point to a general rule (the differential equation) that defines the entire path of the curve.

Q: How do I know for sure that $C=0$?

Great catch! We found the general solution $y = \frac{3}{2x} + Cx$. To find the specific curve, we use the boundary condition provided: the curve passes through $P(1, \frac{3}{2})$. By substituting $x=1$ and $y=\frac{3}{2}$ into our general solution, we get $\frac{3}{2} = \frac{3}{2} + C$. This forces $C$ to be zero. It is not a guess; it is a direct consequence of the curve being anchored at that specific point.

Q: Is there a shortcut? Can I just guess that it's a hyperbola?

While your intuition is a superpower, in JEE Advanced, we must prove it. Guessing might work if you recognize the form $2xy=3$ early, but the rigorous path—deriving the differential equation—is what ensures you don't miss any subtle constraints or alternative solutions. Trust the calculus; it is the most reliable map you have.

Question

Let a curve $y = f(x), x \in (0, \infty)$ pass through the points $P(1, \frac{3}{2})$ and $Q(a, \frac{1}{2})$. If the tangent at any point $R(b, f(b))$ to the given curve cuts the $y$-axis at the point $S(0, c)$ such that $bc = 3$, then $(PQ)^2$ is equal to _____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are walking along a path in the first quadrant, a curve defined by $y = f(x)$. At any point $R(b, f(b))$ on this path, you decide to draw a tangent line.

This line, like a straight arrow shot from the curve, continues until it pierces the $y$-axis at a point $S(0, c)$. The problem gives us a beautiful, simple constraint: the product of the $x$-coordinate of the point of tangency and the $y$-intercept of the tangent is always three, or $bc = 3$.

To begin, we must translate this geometric description into the language of calculus. The slope of the tangent at $R(b, f(b))$ is given by the derivative $f'(b)$.

Using the point-slope form, the equation of this tangent line is $Y - f(b) = f'(b)(X - b)$. We want to find where this line hits the $y$-axis, so we set the capital $X$ coordinate to zero.

This gives us $c - f(b) = f'(b)(0 - b)$, which simplifies to $c = f(b) - b f'(b)$. This expression for $c$ is our key.

### The Bridge to Calculus

Now, we invoke the condition $bc = 3$. Substituting our expression for $c$, we get $b(f(b) - b f'(b)) = 3$.

Since this must hold for any point on the curve, we can generalize this by replacing $b$ with $x$ and $f(b)$ with $y$. We arrive at the differential equation:

$$x \left( y - x \frac{dy}{dx} \right) = 3$$

Let us expand this: $xy - x^2 \frac{dy}{dx} = 3$. To make this look like a standard linear differential equation, we divide by $-x^2$, yielding:

$$\frac{dy}{dx} - \frac{y}{x} = -\frac{3}{x^2}$$

This is the standard form $\frac{dy}{dx} + Py = Q$, where $P = -\frac{1}{x}$ and $Q = -\frac{3}{x^2}$.

### The Integrating Factor: A Magical Tool

We need an Integrating Factor (IF) to unravel the derivative. The IF is defined as $e^{\int P dx}$.

Here, that is $e^{\int -\frac{1}{x} dx}$, which simplifies to $e^{-\ln x}$, or simply $\frac{1}{x}$. Multiplying our entire differential equation by this factor, we get:

$$\frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = -\frac{3}{x^3}$$

Notice the left side? It is the derivative of the product $\frac{y}{x}$ with respect to $x$. Integrating both sides, we get:

$$\frac{y}{x} = \int -\frac{3}{x^3} dx = -3 \int x^{-3} dx = -3 \left( \frac{x^{-2}}{-2} \right) + C = \frac{3}{2x^2} + C$$

### The Final Reveal

We have the family of curves $\frac{y}{x} = \frac{3}{2x^2} + C$. We know the curve passes through $P(1, \frac{3}{2})$.

Substituting these values, we find $\frac{3/2}{1} = \frac{3}{2(1)^2} + C$, which leads to $C = 0$. The equation of our curve is $y = \frac{3}{2x}$, or **$2xy = 3$**.

Finally, we find point $Q(a, \frac{1}{2})$ on this curve. Substituting $x=a$ and $y=\frac{1}{2}$ into $2xy=3$, we get $2(a)(\frac{1}{2}) = 3$, so $a = 3$. Thus, $Q$ is $(3, \frac{1}{2})$.

The square of the distance $PQ$ is:

$$(3 - 1)^2 + \left( \frac{1}{2} - \frac{3}{2} \right)^2 = 2^2 + (-1)^2 = 4 + 1 = 5$$

We have successfully decoded the geometry and arrived at the final answer of **$5$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Equation of Tangent at $R$

Finding the $y$ -intercept $c$

Applying the Condition $b c = 3$

Forming the Linear Differential Equation

Finding the Integrating Factor

Solving the Differential Equation

Finding the Constant $C$

Final Equation of the Curve

Finding the Point $Q$

Calculating $(P Q)^{2}$

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Bridge to Calculus

The Integrating Factor

The Final Reveal

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Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Equation of Tangent at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

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

Applying the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bc=3