Problem 97 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Assertion: The differential equation of all straight lines which are at a constant distance \(p\) from the origin is \(\left(y-x y_{1}\right)^{2}=p^{2}\left(1+y_{1}^{2}\right)\) Reason: The general equation of any straight line which is at a constant distance \(p\) from the origin is \(x \cos \alpha+y \sin \alpha=p .\)

Step-by-Step Solution

Answer

The assertion is true; the given differential equation is correctly derived for such lines.

Step 1: Understand the Assertion

The assertion claims that the differential equation representing all lines at a constant distance \(p\) from the origin is \( (y - xy_1)^2 = p^2(1 + y_1^2) \). We will verify this by deriving the correct differential equation.

Step 2: Start from the General Equation

The reason correctly gives the equation of a straight line at a constant distance \(p\) from the origin as \(x \cos \alpha + y \sin \alpha = p\). This equation represents a family of lines with varying \(\alpha\).

Step 3: Differentiate the General Equation

Differentiate the equation \(x \cos \alpha + y \sin \alpha = p\) with respect to \(x\). The derivative is given by \(\cos \alpha + y_1 \sin \alpha = 0\), where \(y_1 = \frac{dy}{dx}\).

Step 4: Solve for \(\sin \alpha\) and substitute in the original equation

From the differentiated equation \(\cos \alpha + y_1 \sin \alpha = 0\), solve for \(\sin \alpha\): \(\sin \alpha = -\frac{\cos \alpha}{y_1}\). Substitute \(\sin \alpha\) back into the line's equation: \(x \cos \alpha + y \left(-\frac{\cos \alpha}{y_1}\right) = p\). Simplify to get \( y - x y_1 = -p y_1 \).

Step 5: Eliminate \(\cos \alpha\) using trigonometric identity

Squaring both sides of \((y - x y_1) = -p y_1\), we get \((y - x y_1)^2 = p^2 y_1^2\). Then replace \(y_1^2\) using \(1 + y_1^2 = \sec^2 \alpha\). This implies \(y_1^2 = \sec^2 \alpha - 1\). Substitute back to have \((y - x y_1)^2 = p^2 (1 + y_1^2)\), which confirms the given assertion.

Key Concepts

Straight Lines

In mathematics, a straight line is one of the most fundamental concepts in geometry and algebra. In a Cartesian coordinate system, a straight line is defined by a linear equation of the form \( y = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept. This equation signifies that for any point on the line, the change in \( y \) with respect to \( x \) is constant, illustrating the idea of a slope.
Every straight line can also be described in a general form as \( Ax + By + C = 0 \), which encompasses vertical and horizontal lines too. Here, the coefficients \( A \), \( B \), and \( C \) determine the line's position and orientation.
**Key Properties of Straight Lines**

• **Slope**: The incline or steepness, calculated as the rise over the run between two points.
• **Intercepts**: Points where the line crosses the axes, which can be found by setting \( x \) or \( y \) to zero.
• **Parallel and Perpendicular Lines**: Parallel lines have equal slopes \( m_1 = m_2 \), while perpendicular lines have slopes that multiply to \(-1\) (\( m_1 \cdot m_2 = -1 \)).

Understanding these properties allows us to explore how straight lines interact geometrically, setting a base for more complex shapes and figures.

Constant Distance from Origin

Geometrically, the concept of being at a constant distance from the origin is associated with circles or curved lines in a plane. However, when applied to straight lines, it translates to setting a relationship between the line equations and the origin based on distance.
A straight line at a constant distance \( p \) from the origin can be defined by the formula: \( x \cos \alpha + y \sin \alpha = p \). Here, \( \alpha \) is the angle that the perpendicular from the origin makes with the x-axis. This equation represents an infinite set of lines, each maintaining the specified distance from the origin.
**Visualizing Constant Distance**

• **Perpendicular Distance**: For any point \((x_1, y_1)\) on the line, the distance to the origin can be computed using the formula \( \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} = p \). Here, \( A \), \( B \), and \( C \) come from the line equation \( Ax + By + C = 0 \).
• **Implications**: If a line moves such that it always stays at a distance \( p \) from the origin, it forms a path that can be analyzed using differential equations to find its unique properties.

This fundamental understanding aids in exploring how lines maintain equilibrium in a defined space.

Family of Curves

The expression "family of curves" refers to a set of curves that share a common algebraic property or equation but differ by parameter values. When we discuss a family of straight lines, it usually involves a parameter that varies among the lines, such as the angle or intercept.
In the context of differential equations, identifying a family of curves helps establish a deeper understanding of dynamic systems characterized by shared traits.
**Understanding Family of Curves**

• **Parameterization**: Changing a parameter like \( \alpha \) in \( x \cos \alpha + y \sin \alpha = p \) leads to different curves, each maintaining the same distance \( p \) from the origin.
• **Differential Equations**: By deducing a differential equation, as shown above, we can describe the behavior of every curve in the family in relation to others. This approach is crucial in solving more complex differential problems.
• **Applications**: Families are particularly used in physics and engineering to model phenomena where a path or trajectory might take on various forms based on initial conditions or external forces.

Grasping the concept of a family of curves allows students to appreciate the interrelations between different geometric entities and expand their problem-solving toolkit.