Maths · JEE Advanced only

Tangents and Normals to Curves (JEE Advanced)

Not in the JEE Main syllabus, but it is in JEE Advanced. Study time: about 6 to 8 hours. Spend 2 hours on basic tangent and normal equations, 2 hours on lengths (tangent, normal, subtangent, subnormal), 2 hours on parametric and implicit curves, and 2 hours solving mixed problems from past JEE Advanced papers..

Last updated: 26 October 2026

Direct answer

Tangents and normals are lines that touch or cut a curve at a specific point.

Tangents and normals are lines that touch or cut a curve at a specific point. The tangent is a line that just touches the curve at that point and has the same slope as the curve there. The normal is perpendicular to the tangent at that same point.

Official syllabus

What the JEE Advanced syllabus says

“Tangents and normals”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: Applications of derivatives: Rate of change of quantities, monotonic-Increasing and decreasing functions, Maxima and minima of functions of one variable.

JEE Main covers basic derivative applications but does not go deep into geometric properties of curves like tangent lengths, normal lengths, subtangent, and subnormal. For Advanced, you need to handle these concepts along with parametric and implicit curve problems.

Key ideas

What you must understand

The slope of the tangent at point (x₁, y₁) equals dy/dx evaluated at that point
The normal is perpendicular to the tangent, so its slope is the negative reciprocal of the tangent's slope
For parametric curves x = f(t), y = g(t), the slope is (dy/dt)/(dx/dt)
Length of tangent, normal, subtangent, and subnormal are standard results you should memorize
For implicit curves F(x,y) = 0, use implicit differentiation to find dy/dx
Common applications include finding common tangents to two curves and angle of intersection between curves

Formulas

Formulas and key facts

Equation of tangent: y − y₁ = m(x − x₁), where m = dy/dx at (x₁, y₁)
Equation of normal: y − y₁ = (−1/m)(x − x₁)
Length of tangent = |y₁|√(1 + 1/m²) = |y₁√(1 + m²)/m|
Length of normal = |y₁|√(1 + m²)
Length of subtangent = |y₁/m|
Length of subnormal = |y₁ × m|
Angle between two curves: tan θ = |(m₁ − m₂)/(1 + m₁m₂)|
For parametric curves: dy/dx = (dy/dt)/(dx/dt)

Avoid

Common mistakes

Forgetting to use the negative reciprocal for the normal's slope (using m instead of −1/m)
Mixing up the formulas for length of tangent and length of normal
Not substituting the point coordinates back into dy/dx to get the numerical slope
For implicit differentiation, forgetting to apply chain rule when differentiating y terms

Worked example

A JEE Advanced-level example

Question. Find the equation of the normal to the curve y = x³ − 2x² + 4 at the point where x = 2.

Show solution

Find y-coordinate: y = (2)³ − 2(2)² + 4 = 8 − 8 + 4 = 4. So the point is (2, 4).
Find dy/dx: dy/dx = 3x² − 4x
Find slope of tangent at x = 2: m = 3(2)² − 4(2) = 12 − 8 = 4
Slope of normal = −1/m = −1/4
Equation of normal: y − 4 = (−1/4)(x − 2)
Simplify: 4y − 16 = −x + 2, so x + 4y = 18

Answer: x + 4y = 18

Solved twice independently; both solutions agree.

FAQs

Frequently asked questions

Is Tangents and Normals to Curves in the JEE Main syllabus?

No. It is in the JEE Advanced 2026 syllabus only. The closest JEE Main topic is Applications of derivatives: Rate of change of quantities, monotonic-Increasing and decreasing functions, Maxima and minima of functions of one variable.

How long does Tangents and Normals to Curves take to prepare for JEE Advanced?

About 6 to 8 hours. Spend 2 hours on basic tangent and normal equations, 2 hours on lengths (tangent, normal, subtangent, subnormal), 2 hours on parametric and implicit curves, and 2 hours solving mixed problems from past JEE Advanced papers., if you already know the related JEE Main topic.