Tangents and normals: Class 12 practice
The slope of the tangent at x = a is f′(a); the normal is perpendicular, with slope −1/f′(a). Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Differentiation
- Equation of a line
The idea behind tangents and normals
The slope of the tangent at x = a is f′(a); the normal is perpendicular, with slope −1/f′(a). Use point-slope form through (a, f(a)).
Step-by-step method
- Find f(a) and f′(a).
- Tangent: y − f(a) = f′(a)(x − a).
- Normal: slope −1/f′(a).
Worked example 1 (Basic)
Find the equation of the tangent to y = 1x² + 4x at x = -2.
- Point (-2, -4)
- Slope = 2(-2) + 4 = 0
- y = 0x − 4
- Answer: y = 0x − 4
Worked example 2 (Standard)
Find the equation of the tangent to y = 2x² − 4x at x = -1.
- Point (-1, 6)
- Slope = 4(-1) − 4 = -8
- y = -8x − 2
- Answer: y = -8x − 2
Worked example 3 (Challenge)
Find the equation of the normal to y = 2x² + 3x at x = 0.
- Tangent slope 3, normal slope -1/3
- y − 0 = -1/3(x − 0) → x + 3y = 0
- Answer: x + 3y = 0
Common mistakes and misconceptions
- Using f(a) as the slope
- Forgetting the negative reciprocal for the normal
Practice questions
- (Basic) Find the equation of the tangent to y = 3x² + 4x at x = 3.
- (Basic) Find the equation of the tangent to y = 2x² − 2x at x = -1.
- (Standard) Find the equation of the tangent to y = 2x² − 1x at x = -1.
- (Standard) Find the equation of the tangent to y = 1x² − 2x at x = 3.
- (Challenge) Find the equation of the normal to y = 1x² + 3x at x = 2.
- (Challenge) Find the equation of the normal to y = 3x² − 2x at x = 2.
Answer key with full working
- Q1: Point (3, 39) → Slope = 6(3) + 4 = 22 → y = 22x − 27 → Answer: y = 22x − 27
- Q2: Point (-1, 4) → Slope = 4(-1) − 2 = -6 → y = -6x − 2 → Answer: y = -6x − 2
- Q3: Point (-1, 3) → Slope = 4(-1) − 1 = -5 → y = -5x − 2 → Answer: y = -5x − 2
- Q4: Point (3, 3) → Slope = 2(3) − 2 = 4 → y = 4x − 9 → Answer: y = 4x − 9
- Q5: Tangent slope 7, normal slope -1/7 → y − 10 = -1/7(x − 2) → x + 7y = 72 → Answer: x + 7y = 72
- Q6: Tangent slope 10, normal slope -1/10 → y − 8 = -1/10(x − 2) → x + 10y = 82 → Answer: x + 10y = 82
Syllabus note
This skill is listed under "Application of Derivatives" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is tangents and normals in?
"Application of Derivatives", Class 12 Maths (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.