Tangents and Normals to Curves: Mastering JEE Calculus

In JEE Mathematics, understanding tangents and normals to curves is crucial for solving problems involving rates of change, optimization, and geometric properties of functions. This module will equip you with the fundamental concepts and techniques to confidently tackle these questions.

What are Tangents and Normals?

A tangent to a curve at a point is a straight line that touches the curve at that single point and has the same slope as the curve at that point. Imagine a wheel rolling on a flat surface; the point where the wheel touches the surface at any instant is like a tangent.

A normal to a curve at a point is a straight line that is perpendicular to the tangent line at that same point. It represents the direction perpendicular to the curve's instantaneous direction of motion.

The Role of the Derivative

The derivative of a function, $f'(x)$ , at a specific point $x=a$ gives the slope of the tangent line to the curve $y=f(x)$ at that point $(a, f(a))$ . This is the cornerstone of finding tangents and normals.

Finding the Equation of the Tangent Line

To find the equation of the tangent line at a point $(x_1, y_1)$ on the curve $y=f(x)$ :

Calculate the derivative $f'(x)$ .
Evaluate the derivative at $x=x_1$ to find the slope of the tangent, $m_t = f'(x_1)$ .
Use the point-slope form of a line: $y - y_1 = m_t(x - x_1)$ .

Finding the Equation of the Normal Line

Since the normal line is perpendicular to the tangent line, their slopes are negative reciprocals of each other. If the slope of the tangent is $m_t$ , the slope of the normal ( $m_n$ ) is given by $m_n = -1/m_t$ (provided $m_t eq 0$ ).

To find the equation of the normal line at a point $(x_1, y_1)$ on the curve $y=f(x)$ :

Find the slope of the tangent line, $m_t = f'(x_1)$ .
Calculate the slope of the normal line, $m_n = -1/m_t$ (if $m_t=0$ , the tangent is horizontal, and the normal is vertical with equation $x=x_1$ ).
Use the point-slope form of a line: $y - y_1 = m_n(x - x_1)$ .

Remember: If the tangent is horizontal ( $m_t=0$ ), the normal is vertical ( $x=x_1$ ). If the tangent is vertical ( $m_t$ is undefined), the normal is horizontal ( $y=y_1$ ).

Special Cases and Applications

Problems might involve finding tangents or normals that are parallel or perpendicular to a given line, or finding points where the tangent/normal has a specific slope. These often require setting up equations based on the slope conditions.

Consider a curve $y = x^2$ . We want to find the tangent and normal at the point (2, 4).

Find the derivative: $y' = \frac{d}{dx}(x^2) = 2x$ .
Slope of the tangent ( $m_t$ ) at x=2: $m_t = 2(2) = 4$ .
Equation of the tangent: Using point-slope form $y - y_1 = m_t(x - x_1)$ , we get $y - 4 = 4(x - 2)$ , which simplifies to $y = 4x - 4$ .
Slope of the normal ( $m_n$ ): $m_n = -\frac{1}{m_t} = -\frac{1}{4}$ .
Equation of the normal: Using point-slope form $y - y_1 = m_n(x - x_1)$ , we get $y - 4 = -\frac{1}{4}(x - 2)$ , which simplifies to $4y - 16 = -x + 2$ , or $x + 4y = 18$ .

📚

Text-based content

Library pages focus on text content

What is the relationship between the slope of a tangent line and the derivative of the function at that point?

The derivative of the function at a point is equal to the slope of the tangent line to the curve at that point.

How is the slope of the normal line related to the slope of the tangent line?

The slope of the normal line is the negative reciprocal of the slope of the tangent line (i.e., $m_n = -1/m_t$ ).

Practice Problems for JEE

JEE problems often involve:

Finding the equation of tangent/normal to parametric curves or implicit functions.
Determining points on a curve where the tangent is parallel or perpendicular to a given line.
Calculating the angle between two curves (which involves the angle between their tangents).
Problems related to subtangent, subnormal, length of tangent, and length of normal.

Key Formulas to Remember

Concept	Formula
Slope of Tangent at $(x_1, y_1)$	$m_t = f'(x_1)$
Equation of Tangent at $(x_1, y_1)$	$y - y_1 = m_t(x - x_1)$
Slope of Normal at $(x_1, y_1)$	$m_n = -1/m_t$ (if $m_t \neq 0$ )
Equation of Normal at $(x_1, y_1)$	$y - y_1 = m_n(x - x_1)$
Vertical Normal (when $m_t = 0$ )	$x = x_1$

Learning Resources

Tangents and Normals - Introduction and Formulas(documentation)

Provides a clear explanation of tangents and normals, along with essential formulas and solved examples relevant to competitive exams.

Differentiation - Tangents and Normals to Curves(documentation)

This resource offers a comprehensive overview of tangents and normals, including their derivation and application in calculus problems.

Calculus: Tangents and Normals - Khan Academy(video)

A video tutorial explaining the concepts of tangent and normal lines using derivatives, with clear visual aids.

JEE Mathematics: Tangents and Normals(blog)

A collection of questions and answers on StackExchange related to tangents and normals, offering diverse problem-solving approaches.

NCERT Class 12 Maths Chapter 6 - Application of Derivatives(documentation)

The official NCERT textbook chapter on Applications of Derivatives, which includes a detailed section on tangents and normals.

Understanding Tangents and Normals with Examples(documentation)

This guide provides step-by-step examples for finding tangents and normals to various types of curves.

Calculus - Tangents and Normals(documentation)

A simplified explanation of tangents and normals, focusing on the core concepts and their geometric interpretation.

JEE Advanced Mathematics - Tangents and Normals(video)

A video solution to a typical JEE problem involving finding the tangent to a curve, demonstrating practical application.

Applications of Derivatives - Tangents and Normals(documentation)

Detailed notes and solved problems on tangents and normals, specifically tailored for JEE preparation.

Geometric Interpretation of the Derivative(documentation)

Explains the fundamental link between the derivative and the slope of the tangent line, providing a strong theoretical foundation.