Problem 47 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The equation of the family of curves which intersect the hyperbola \(x y=2\) orthogonally is (A) \(y=\frac{x^{3}}{6}+C\) (B) \(y=\frac{x^{2}}{4}+C\) (C) \(y=-\frac{x^{3}}{6}+C\) (D) \(y=-\frac{x^{2}}{4}+C\)

Step-by-Step Solution

Answer

(A) \( y = \frac{x^3}{6} + C \).

Step 1: Determine the Slope of the Hyperbola

Given the hyperbola equation is \( x y = 2 \). First, differentiate both sides with respect to \( x \) using implicit differentiation. We get\[dy/dx = -2/x^2\]This is the slope of the tangent to the hyperbola at any point \( (x, y) \).

Step 2: Establish Orthogonality Condition

Curves that are orthogonal to the hyperbola have tangents that satisfy the negative reciprocal condition. Therefore, if the slope of the hyperbola is \( -2/x^2 \), the orthogonal slope is \[y' = x^2/2\]

Step 3: Solve the Differential Equation

Integrate the orthogonal slope equation to find the family of curves. We have:\[y' = \frac{x^2}{2}\]Integrate with respect to \( x \):\[y = \int \frac{x^2}{2} \, dx = \frac{x^3}{6} + C\]where \( C \) is the constant of integration.

Step 4: Match to Given Options

Compare the derived equation \( y = \frac{x^3}{6} + C \) with the given options. It matches option (A).

Key Concepts

Understanding Hyperbola

A hyperbola is a type of smooth curve lying in a plane, defined as the set of all points such that the difference of the distances to two fixed points (the foci) is constant. In this exercise, the specific hyperbola given is represented by the equation \( xy = 2 \).

• This form characterizes a rectangular hyperbola due to its symmetrical structure across the axes.
• Understanding the slope of a tangent line to the hyperbola at any given point is essential for further solving problems involving orthogonal intersections.

The slope is key because it helps to determine how other curves may intersect or interact with the hyperbola. The slope is calculated via differentiation, leading us into exploring implicit differentiation, which is crucial for this problem.

Implicit Differentiation Simplified

Implicit differentiation is a technique used to find the derivative of a function that is not isolated on one side of an equation. This is particularly useful when dealing with equations like hyperbola \( xy = 2 \).By differentiating both sides of the hyperbola's equation with respect to \( x \), we reveal the slope of the tangent line to the hyperbola at any point \( (x,y) \). Here, implicit differentiation involves:

• Applying the product rule to \( xy \) which results in \( y + x\frac{dy}{dx} = 0 \).
• Solving this for \( \frac{dy}{dx} \), yielding \( \frac{dy}{dx} = -\frac{2}{x^2} \).

This result is integral in identifying how any other curve, particularly those intersecting orthogonally, may interact with the hyperbola. The concept of orthogonality means these intersecting curves must have slopes that are negative reciprocals of the hyperbola's slope.

Differential Equations and Orthogonal Intersections

Differential equations come into play prominently when dealing with orthogonal intersections. To solve for a curve that intersects the hyperbola \( xy = 2 \) orthogonally, we leverage the condition that the product of the slopes of two orthogonal curves at their intersection point is \(-1\).Given the tangent slope of the hyperbola is \( -\frac{2}{x^2} \), the orthogonal slope is found to be \( \frac{x^2}{2} \). This leads to a differential equation for the family of orthogonal curves:

• \( \frac{dy}{dx} = \frac{x^2}{2} \)

Integrating this equation with respect to \( x \) gives:

• \( y = \int \frac{x^2}{2} \, dx = \frac{x^3}{6} + C \)

This integration process reveals the family of curves that satisfy the orthogonality condition, matching option (A) from the given choices. Thus, differential equations enable us to explore and define complex systems and interactions like orthogonal intersections analytically and effectively.