Problem 84 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The lines \(p\left(p^{2}+1\right) x-y+q=0\) and \(\left(\mathrm{p}^{2}+1\right)^{2} \mathrm{x}+\left(\mathrm{p}^{2}+1\right) \mathrm{y}+2 \mathrm{q}=0\) are perpendicular to a common line for : \(\quad\) [2009] (a) exactly one values of \(\mathrm{p}\) (b) exactly two values of \(\mathrm{p}\) (c) more than two values of \(\mathrm{p}\) (d) no value of \(\mathrm{p}\)

Step-by-Step Solution

Answer

Only \( p = 1 \) is solution; exactly one value of \( p \).

Step 1: Understand the Geometry

Two lines are perpendicular to a common line if their slopes are negative reciprocals of each other. To find the slopes of the given lines, we need to identify the coefficients of x and y in each equation and express them in the slope-intercept form.

Step 2: Find the Slope of the First Line

The first line is given by the equation \( p(p^2 + 1)x - y + q = 0 \). Isolation of \( y \) gives us \( y = p(p^2 + 1)x + q \). Thus, the slope of this line is \( m_1 = p(p^2 + 1) \).

Step 3: Find the Slope of the Second Line

The second line is \( (p^2 + 1)^2 x + (p^2 + 1) y + 2q = 0 \). Simplifying to the slope-intercept form gives \( y = -\frac{(p^2 + 1)^2}{(p^2 + 1)} x - \frac{2q}{p^2 + 1} \), so the slope is \( m_2 = -(p^2 + 1) \).

Step 4: Set Slopes as Negative Reciprocals

For the two lines to be perpendicular to a common line, we require that the product of their slopes must be \( -1 \). Set \( m_1 \cdot m_2 = -1 \) which simplifies to \( p(p^2 + 1) \cdot (-(p^2 + 1)) = -1 \).

Step 5: Simplify the Equation

We have \( -p(p^2 + 1)^2 = -1 \). This simplifies to \( p(p^2 + 1)^2 = 1 \). Thus, \( p(p^4 + 2p^2 + 1) = 1 \).

Step 6: Solve for \( p \)

Expand and simplify the equation as \( p^5 + 2p^3 + p = 1 \). This is a polynomial equation that might have multiple roots. One known strategy is to test practical roots or assert its polynomial nature formulaically.

Step 7: Analyze Roots Potential

Given the degree of the polynomial (5th degree), the possible number of solutions or roots for \( p \) could be up to 5, considering both real and complex roots. However, direct calculation or factoring shows that \( p = 1 \) satisfy the equation.

Key Concepts

Understanding Geometry

Geometry is an essential branch of mathematics, dealing with shapes, sizes, and the properties of space. In this problem, geometry helps us understand how two lines relate to each other when they are perpendicular to a common line.
In simpler terms, two lines are perpendicular to the same line if their angles with this line are complementary, which means their slopes must be negative reciprocals.

• Imagine two lines meeting at right angles to another line in space.
• Visualizing this with real-world objects, like streets intersecting a central plaza, can help solidify the concept.

An equation of a line typically helps in finding such geometric relationships when expressed in forms like slope-intercept.

Slope-Intercept Form

The slope-intercept form of a line serves as a fundamental format in mathematics. It's written as:
\( y = mx + c \), where

• \( m \) represents the slope, and
• \( c \) is the y-intercept of the line.

The slope \( m \) indicates how steep the line is. The bigger the slope, the steeper the line.
Understanding this form is crucial because it directly tells us the line's direction and how it intersects the y-axis.
In this exercise, transforming the given equations into the slope-intercept form allowed us to easily identify the slopes of the two lines, a critical step in solving the problem.

Characterizing Perpendicular Lines

Perpendicular lines are lines that intersect each other at right angles, or 90 degrees.
If two lines are perpendicular, their slopes multiply together to give
\(-1\).
In mathematical terms, if line 1 has a slope of \(m_1\) and line 2 has a slope of \(m_2\), then

• \( m_1 \times m_2 = -1 \)

This concept was crucial in our specific exercise, where the slopes needed to be negative reciprocals to satisfy perpendicularity to a common line.
Understanding such relationships about perpendicularity extends beyond this exercise and applies to many geometric and real-world problems.

Solving Polynomial Equations

Polynomial equations are mathematical expressions consisting of variables and coefficients, involving terms in the form of a variable raised to a power. The challenge often lies in finding values of the variable that make the equation true.
Our exercise led to a 5th-degree polynomial:
\( p^5 + 2p^3 + p = 1 \).

• This indicates a potentially complex equation requiring methodical approaches.

Examples of such approaches include factoring, synthetics division, or using the Rational Root Theorem, which involves testing roots like \(p = 1\).
Working with polynomials hones your ability to navigate complex algebraic expressions and is a key skill to master for advancing in math.