Problem 30 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Given that the slope of the tangent to a curve \(y=y(x)\) at any point \((x, y)\) is \(\frac{2 y}{x^{2}}\). If the curve passes through the centre of the circle \(x^{2}+y^{2}-2 x-2 y=0\), then its equation is: (a) \(x \log _{e}|y|=2(x-1)\) (b) \(x \log _{e}|y|=-2(x-1)\) (c) \(x^{2} \log _{e}|y|=-2(x-1)\) (d) \(x \log _{e}|y|=x-1\)

Step-by-Step Solution

Answer

The correct equation is (b) \(x \log_{e}|y| = -2(x-1)\).

Step 1: Identify the center of the circle

Rewrite the equation of the circle as \((x-1)^2 + (y-1)^2 = 2\) by completing the square. From this, identify the center of the circle as \((1, 1)\).

Step 2: Use the initial condition

Since the curve passes through \((1, 1)\), any particular solution to the differential equation must satisfy \(y(1) = 1\).

Step 3: Solve the differential equation

We have \(\frac{dy}{dx} = \frac{2y}{x^2}\). This can be rewritten as \(\frac{dy}{y} = \frac{2}{x^2}dx\). Integrate both sides to get \(\ln|y| = -\frac{2}{x} + C\), where \(C\) is the integration constant.

Step 4: Apply the initial condition

Substitute \((x, y) = (1, 1)\) into \(\ln|y| = -\frac{2}{x} + C\). This gives \(\ln(1) = -2 + C\), thus \(C = 2\).

Step 5: Express the general solution

The solution to the differential equation is \(\ln|y| = -\frac{2}{x} + 2\). This can also be expressed as \(y = e^{\left(-\frac{2}{x} + 2\right)} = e^2 e^{-\frac{2}{x}}\).

Step 6: Rearrange the equation to match options

From \(\ln|y| = -\frac{2}{x} + 2\), we can rearrange and simplify the expression \(x \ln |y| = -2(x - 1)\).

Key Concepts

Understanding the Slope of Tangent

The slope of the tangent line to a curve at any given point is a fundamental concept in calculus. It's a way of describing how steep the curve is at that particular spot. Here, the problem gives the slope of the tangent as \( \frac{2y}{x^2} \). In simple terms:

• The slope tells us the direction in which the curve is heading.
• Mathematically, it corresponds to the derivative of \( y \) with respect to \( x \), denoted as \( \frac{dy}{dx} \).
• This slope formula influences the shape and direction of the curve.

In this exercise, understanding the slope allows us to form a differential equation. This equation will later be used to find the actual function that \( y \) represents.

Integration: Solving the Slope Equation

Integration is the mathematical process used to solve differential equations. Here, we're given \( \frac{dy}{dx} = \frac{2y}{x^2} \) and are tasked with finding \( y(x) \). Integrating both sides helps us achieve this:

• Divide both sides to separate variables: \( \frac{dy}{y} = \frac{2}{x^2}dx \).
• Separate integration gives us \( \int \frac{dy}{y} = \int \frac{2}{x^2} dx \).
• On integration, we get \( \ln|y| = -\frac{2}{x} + C \), where \( C \) is a constant of integration.

Integration helps convert the derivative (slope) into a function that describes the original curve. This allows us to use known conditions to solve for \( y \).

Applying Initial Conditions

Initial conditions are central to determining unique solutions from general ones. In this problem, the curve is known to pass through the point \((1, 1)\). Here's how we employ this:

• We substitute \( (x, y) = (1, 1) \) into \( \ln|y| = -\frac{2}{x} + C \).
• This gives us \( \ln(1) = -2 + C \), and since \( \ln(1) = 0 \), it follows that \( C = 2 \).
• Using the initial condition ensures the solution fits this particular case, tailoring the general solution to match specific points on the curve.

Using such conditions narrows our solution down to match the problem’s specific scenario.

Solving the Differential Equation

The overall process of solving a differential equation combines solving the slope equation and applying initial conditions.

• Start by integrating the differential equation to find a general solution.
• Apply any initial conditions or known points to find specific constants of integration.
• For this exercise, the steps lead us to \( \ln|y| = -\frac{2}{x} + 2 \).
• Rearrange to match one of the provided options: \( x \ln |y| = -2(x - 1) \).

This process results in a specific, applicable function that represents our curve, given the slope information and initial conditions.