Tangents & Normals to Curves
Applications of Derivatives
40
JEE Qs
10%
Hard
75
min
Master all differentiation techniques and the exact geometric interpretation of dy/dx at a point to solve complex problems efficiently.
🧮 Key Formulas
✅ Key Points for JEE
- 1Always find the point of tangency (x₀, y₀) first. If not given, assume it or find it using given conditions.
- 2Be proficient in all differentiation techniques (implicit, parametric, chain rule) to correctly find dy/dx.
- 3If the tangent is parallel to the x-axis, dy/dx = 0. If parallel to the y-axis, dx/dy = 0 (or dy/dx is undefined).
- 4For the angle between two curves, find their slopes at each point of intersection. Multiple intersection points might yield different angles.
- 5A common tangent to two curves implies that at the point of tangency, the coordinates and the slopes (dy/dx) are equal for both curves.
⚠️ Common Mistakes
- ✕Incorrectly calculating dy/dx, especially for implicit or parametric functions, leading to wrong slopes.
- ✕Confusing the slope of the tangent with the slope of the normal, or using the reciprocal without the negative sign.
- ✕Making algebraic errors when solving for the point of tangency or intersection.
- ✕Not checking for special cases like vertical tangents (dy/dx undefined, meaning dx/dy = 0).
📝 Practice Questions
See allQ10.Let the function f(x) = (x2 + 1) x2 −ax + 2 + cos |x| be not differentiable at the two points x = α = 2 and x = β . Then the distance of the point (α, β) from the line 12x + 5y + 10 = 0 is equal to : (1) 5 (2) 4 (3) 3 (4) 2
Q24.Let the function, f(x) = {−3ax2a2 + bx,−2, xx <⩾11 be differentiable for all x ∈R, where a > 1, b ∈R. If the area of the region enclosed by y = f(x) and the line y = −20 is α + β√3, α, β ∈Z , then the value of α + β is ________
Q8. Let f(x) = ∫x20 t2−8t+15et dt, respectively, are : (1) 2 and 3 (2) 2 and 2 (3) 3 and 2 (4) 1 and 3
Q23.If the set of all values of a, for which the equation 5x3 −15x −a = 0 has three distinct real roots, is the interval (α, β), then β −2α is equal to ______
Q13.A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm , the ice-cream melts at the rate of 81 cm3/min and the thickness of the ice-cream layer decreases at the rate of 1 cm/min. The surface area (in cm2 ) of the chocolate ball (without the ice- 4π cream layer) is : (1) 196π (2) 256π (3) 225π (4) 128π
Q20.Let →a = ^i + 2^j + 3^k,→b = 3^i + ^j −^k and →c be three vectors such that →c is coplanar with →a and →b. If the vector →C is perpendicular to →b and →a ⋅→c = 5, then |→c| is equal to (1) √116 (2) 3√21 (3) 16 (4) 18
NCERT Chapters
- Class 11 Maths Ch 10: Straight Lines
- Class 12 Maths Ch 5: Continuity and Differentiability
- Class 12 Maths Ch 6: Applications of Derivatives