JEE Tangents & Normals: Top 10 Problem Types Demystified
Your ultimate revision guide for mastering the most frequent and challenging tangent & normal problems in JEE Main & Advanced.
Why Tangents & Normals Are JEE Favorites
Tangents and Normals consistently appear in 8-12% of JEE Calculus questions. These problems test your understanding of derivatives, coordinate geometry, and application skills simultaneously.
🎯 JEE Examination Pattern
- JEE Main: Usually 1-2 questions per paper (4-8 marks)
- JEE Advanced: Often combined with other concepts in multi-step problems
- Most Common: Finding equations, angles, lengths, and intersection points
📐 Essential Formulas - Quick Recall
Tangent at point $(x_1, y_1)$:
$y - y_1 = \left(\frac{dy}{dx}\right)_{(x_1,y_1)} (x - x_1)$
Normal at point $(x_1, y_1)$:
$y - y_1 = -\frac{1}{\left(\frac{dy}{dx}\right)_{(x_1,y_1)}} (x - x_1)$
Angle between curves:
$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|$
Length of tangent:
$PT = \sqrt{y_1^2 + \left(\frac{y_1}{m}\right)^2}$
Type 1: Basic Tangent Equation
Find the equation of tangent to the curve $y = x^3 - 3x^2 + 2x$ at point where $x = 1$.
Solution Approach:
Step 1: Find point: When $x = 1$, $y = 1 - 3 + 2 = 0$ → Point: $(1, 0)$
Step 2: Find derivative: $\frac{dy}{dx} = 3x^2 - 6x + 2$
Step 3: Slope at $x = 1$: $m = 3(1)^2 - 6(1) + 2 = -1$
Step 4: Tangent equation: $y - 0 = -1(x - 1)$
Final Answer: $x + y - 1 = 0$
💡 Quick Recall Formula
Tangent: $y - y_1 = m(x - x_1)$ where $m = f'(x_1)$
Type 2: Tangent Parallel to Given Line
Find points on curve $y = x^3 - 3x$ where tangent is parallel to line $y = 3x - 1$.
Solution Approach:
Step 1: Slope of given line: $m = 3$
Step 2: Derivative: $\frac{dy}{dx} = 3x^2 - 3$
Step 3: Set equal to slope: $3x^2 - 3 = 3$
Step 4: Solve: $3x^2 = 6 \Rightarrow x^2 = 2 \Rightarrow x = \pm\sqrt{2}$
Step 5: Find points:
• $x = \sqrt{2}$: $y = 2\sqrt{2} - 3\sqrt{2} = -\sqrt{2}$
• $x = -\sqrt{2}$: $y = -2\sqrt{2} + 3\sqrt{2} = \sqrt{2}$
Final Answer: $(\sqrt{2}, -\sqrt{2})$ and $(-\sqrt{2}, \sqrt{2})$
💡 Quick Recall Formula
Parallel lines have equal slopes: $f'(x) = m_{\text{line}}$
Type 3: Angle Between Two Curves
Find the angle between curves $y^2 = 4x$ and $x^2 = 4y$ at their point of intersection in first quadrant.
Solution Approach:
Step 1: Find intersection point:
From $y^2 = 4x \Rightarrow x = \frac{y^2}{4}$
Substitute in $x^2 = 4y$: $\left(\frac{y^2}{4}\right)^2 = 4y$
$\frac{y^4}{16} = 4y \Rightarrow y^4 = 64y \Rightarrow y(y^3 - 64) = 0$
$y = 0$ or $y = 4$ (first quadrant)
When $y = 4$, $x = \frac{16}{4} = 4$ → Point: $(4, 4)$
Step 2: Find slopes:
Curve 1: $y^2 = 4x \Rightarrow 2y\frac{dy}{dx} = 4 \Rightarrow m_1 = \frac{2}{y} = \frac{1}{2}$
Curve 2: $x^2 = 4y \Rightarrow 2x = 4\frac{dy}{dx} \Rightarrow m_2 = \frac{x}{2} = 2$
Step 3: Angle between curves:
$\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right| = \left|\frac{\frac{1}{2} - 2}{1 + 1}\right| = \left|\frac{-\frac{3}{2}}{2}\right| = \frac{3}{4}$
Final Answer: $\theta = \tan^{-1}\left(\frac{3}{4}\right)$
💡 Quick Recall Formula
Angle between curves: $\tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right|$
🚀 Time-Saving Strategies
For Tangents:
- Always find both point and slope first
- For implicit curves, use implicit differentiation
- Check if tangent passes through given external point
- Remember perpendicular to normal, parallel to tangent
Common Pitfalls:
- Forgetting to find y-coordinate from x-coordinate
- Mixing up tangent and normal equations
- Not considering multiple intersection points
- Missing absolute value in angle formula
Type 4: Length of Tangent/Normal
For curve $y = x^2 + 2x + 1$, find length of tangent and normal at point $(1, 4)$.
Solution Approach:
Step 1: Find derivative: $\frac{dy}{dx} = 2x + 2$
Step 2: Slope at $(1, 4)$: $m = 2(1) + 2 = 4$
Step 3: Length of tangent = $\frac{|y_1|\sqrt{1 + m^2}}{|m|} = \frac{4\sqrt{1 + 16}}{4} = \sqrt{17}$
Step 4: Length of normal = $|y_1|\sqrt{1 + m^2} = 4\sqrt{17}$
Final Answer: Tangent length = $\sqrt{17}$, Normal length = $4\sqrt{17}$
💡 Quick Recall Formulas
Length of tangent = $\frac{|y_1|\sqrt{1 + m^2}}{|m|}$
Length of normal = $|y_1|\sqrt{1 + m^2}$
Type 5: Common Tangents to Two Curves
Find the equation of common tangent to curves $y = x^2$ and $y = -x^2 + 4x - 4$.
Solution Approach:
Step 1: Let tangent to $y = x^2$ be $y = mx + c$
Step 2: Condition for tangency: discriminant = 0
$x^2 = mx + c \Rightarrow x^2 - mx - c = 0$
Discriminant: $m^2 + 4c = 0 \Rightarrow c = -\frac{m^2}{4}$
Step 3: This line also touches $y = -x^2 + 4x - 4$
$-x^2 + 4x - 4 = mx - \frac{m^2}{4}$
$-x^2 + (4 - m)x - 4 + \frac{m^2}{4} = 0$
Step 4: Discriminant = 0 for tangency:
$(4 - m)^2 - 4(-1)\left(-4 + \frac{m^2}{4}\right) = 0$
$16 - 8m + m^2 - 16 + m^2 = 0$
$2m^2 - 8m = 0 \Rightarrow 2m(m - 4) = 0$
Step 5: For $m = 0$: $y = 0$ (x-axis)
For $m = 4$: $y = 4x - 4$
Final Answer: $y = 0$ and $y = 4x - 4$
Other Essential Problem Types
Types 6-10:
- Type 6: Tangents from external point
- Type 7: Normal with given conditions
- Type 8: Subtangent and subnormal
- Type 9: Parametric curves
- Type 10: Combined with coordinate geometry
JEE Frequency:
- High Frequency: Types 1, 2, 3, 4
- Medium Frequency: Types 5, 6, 7
- Advanced Only: Types 8, 9, 10
📝 Quick Self-Test
Try these problems to test your understanding:
1. Find tangent to $y = x^3 - x$ parallel to x-axis
2. Angle between $y = \sin x$ and $y = \cos x$ at intersection
3. Length of normal to $y^2 = 4ax$ at $(at^2, 2at)$
✅ Final Revision Checklist
Must-Know Formulas:
- Tangent & Normal equations
- Angle between curves formula
- Length of tangent/normal
- Condition for common tangents
- Subtangent & subnormal formulas
Problem Solving Steps:
- Identify curve type and given conditions
- Find derivative and slope
- Locate point of contact
- Apply appropriate formula
- Verify with quick check
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