Use the local direction of a curve without losing vertical lines, singular points or the point of contact.
JEE Main · JEE Advanced60 min study guide9 indexed sections
BY THE END, YOU CAN
Derive tangent and normal equations from slopes or direction vectors.
Handle explicit, implicit and parametric curves with the correct regularity conditions.
Find contacts when a tangent must pass through a prescribed point.
Distinguish touching, crossing, perpendicularity and singular behavior.
BEFORE YOU START
Derivatives, chain rule and implicit differentiation.
Point-slope and general equations of a line.
Dot products of two-dimensional vectors.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. A tangent describes local direction
A secant line joins two points of a curve. If one point approaches the other and the secant directions approach a unique limiting line, that line is a tangent. For a differentiable graph y = f(x), its finite slope at x = a is f′(a). The tangent passes through the actual curve point (a,f(a)); the derivative provides its direction, not its position.
The everyday phrase “a line that touches but does not cross” is unreliable. The tangent y = 0 to y = x³ at the origin crosses the curve there. A tangent can also meet the same curve again somewhere else. Tangency is a local statement about direction near a specified contact, not a global prohibition on intersections.
A normal is a line through the contact perpendicular to the tangent. When the tangent has a finite nonzero slope m, the normal slope is −1/m. If m = 0, however, the tangent is horizontal and the normal is vertical. Writing −1/0 is not a valid slope calculation. Use the geometric direction instead.
Before differentiating, verify that the proposed contact lies on the curve and in its domain. A line calculated at a forbidden x-value is not a tangent to the original graph even if a simplified expression has a removable extension there. The original domain determines which points actually exist.
02 / UNDERSTAND & APPLY
2. Tangent and normal formulas without exceptions hidden
For y = f(x) differentiable at a with m = f′(a), the tangent is y − f(a) = m(x − a). This formula remains valid at m = 0. A useful normal equation that also works when m = 0 is x − a + m(y − f(a)) = 0. When m is nonzero it rearranges to the familiar negative-reciprocal slope formula; when m is zero it becomes x = a.
The vector viewpoint explains both equations. The tangent direction is (1,m). A normal direction is (−m,1), whose dot product with (1,m) is zero. The normal line equation says that the displacement from the contact is perpendicular to the tangent direction. This avoids unnecessary division and makes horizontal cases routine.
These formulas assume the graph has a finite derivative. A vertical tangent, such as at the origin of y = x^(1/3), has no finite slope. Its tangent equation is x = 0 and its normal is y = 0. One can establish this with a limiting secant direction or by writing the same curve as x = y³ and differentiating with respect to y. An undefined derivative alone is not enough: a corner may have two different one-sided directions.
Tangent: y − f(a) = f′(a)(x − a)
Use when the derivative at a exists and is finite.
Normal: x − a + f′(a)(y − f(a)) = 0
This division-free form includes the vertical normal at a horizontal tangent.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Equal x/y scale; the tangent and normal are perpendicular at the common contact.
ORIGINAL WORKED EXAMPLE
A polynomial contact
Find the tangent and normal to y = x² + x at x = 1.
The contact point is (1,2), not (1,1). Differentiate to get y′ = 2x + 1, so the tangent slope is 3.
The tangent equation is y − 2 = 3(x − 1), hence y = 3x − 1.
The normal equation is x − 1 + 3(y − 2) = 0, hence x + 3y − 7 = 0.
Check both equations at (1,2). Their slopes 3 and −1/3 multiply to −1, confirming perpendicularity.
Answer Tangent y = 3x − 1; normal x + 3y − 7 = 0.
03 / UNDERSTAND & APPLY
3. A line through a point may touch elsewhere
When the problem gives a point through which a tangent passes, do not assume it is the contact. Introduce an unknown contact parameter a and write the tangent there. Substitute the given point into that line equation. The resulting equation determines possible contacts, which must then be checked in the original curve and line conditions.
This process can yield zero, one, or several tangents. A point on a curve may have additional tangents touching at other points. An external point may have more than two tangents to a general curve; the familiar two-tangent picture is specific to suitable conics and positions. Always solve the actual contact equation rather than relying on the circle diagram.
When the algebra squares an expression or clears a denominator, preserve restrictions and check candidate contacts afterward. Distinct contact parameters can occasionally produce the same tangent line, a bitangent. If the task asks for distinct lines, compare their equations instead of merely counting roots of the contact equation.
ORIGINAL WORKED EXAMPLE
Two tangents through an external point
Find all tangents to y = x² that pass through (0,−4).
At contact (a,a²), the derivative is 2a and the tangent is y − a² = 2a(x − a), or y = 2ax − a².
To pass through (0,−4), the line must satisfy −4 = −a². Therefore a = 2 or a = −2.
The two tangent equations are y = 4x − 4 and y = −4x − 4. Their contacts are (2,4) and (−2,4).
As an independent check, substituting the first line into y = x² gives (x − 2)² = 0, and substituting the second gives (x + 2)² = 0. Each has the required double contact.
Answer y = 4x − 4 and y = −4x − 4.
04 / UNDERSTAND & APPLY
4. Implicit curves and regular points
For a curve F(x,y) = 0 with differentiable F, implicit differentiation gives Fₓ + Fᵧ y′ = 0 wherever the curve can locally be treated as a differentiable graph y(x) and Fᵧ is nonzero. Thus y′ = −Fₓ/Fᵧ. A safer tangent equation at a regular point (a,b) is Fₓ(a,b)(x − a) + Fᵧ(a,b)(y − b) = 0.
The regular-point condition means the gradient (Fₓ,Fᵧ) is not the zero vector. This vector is perpendicular to the tangent, so it gives the normal direction directly. If Fᵧ = 0 but Fₓ ≠ 0, the tangent equation reduces to x = a and is vertical. The gradient approach handles that case without inventing an infinite numerical slope.
If both partial derivatives vanish, the displayed linear equation becomes 0 = 0 and gives no direction. Such a point requires a separate local analysis. The curve y² = x² has two crossing branches at the origin, with tangents y = x and y = −x. By contrast, y² = x³ has a cusp and a single tangent line y = 0. The zero gradient alone does not tell you which situation occurs.
The choice of defining equation matters at singular-looking points. Squaring a perfectly regular equation, as in F = (x² + y² − 1)², makes its gradient vanish all along the circle. That does not make the circle geometrically singular. Simplify the defining relation when possible and use a representation that retains the local direction information.
Fₓ(a,b)(x − a) + Fᵧ(a,b)(y − b) = 0
At a regular point of F = 0, this is the tangent. The gradient gives the normal direction.
ORIGINAL WORKED EXAMPLE
An ellipse without slope shortcuts
Find the tangent and normal to x² + 4y² = 20 at (2,2).
The point satisfies 4 + 16 = 20. With F = x² + 4y² − 20, the gradient at the point is (4,16), which is nonzero.
The tangent is 4(x − 2) + 16(y − 2) = 0, giving x + 4y = 10. Its slope is −1/4.
The normal has direction (4,16), hence slope 4, and equation y − 2 = 4(x − 2), or y = 4x − 6.
The derivative check gives 2x + 8y y′ = 0, so y′ = −x/(4y) = −1/4 at the point. Both methods agree.
Answer Tangent x + 4y = 10; normal y = 4x − 6.
05 / UNDERSTAND & APPLY
5. Parametric curves: follow the velocity vector
For x = x(t), y = y(t), the vector (x′(t),y′(t)) points along the motion. At a regular parameter value t₀, meaning this vector is nonzero, it supplies a tangent direction even when the curve is not globally a graph. If x′(t₀) ≠ 0, the slope is dy/dx = y′(t₀)/x′(t₀). If x′ = 0 but y′ ≠ 0, the tangent is vertical.
A division-free tangent equation is x′(t₀)(y − y₀) − y′(t₀)(x − x₀) = 0. The normal equation is x′(t₀)(x − x₀) + y′(t₀)(y − y₀) = 0. These equations use the cross-product-zero and dot-product-zero conditions in the plane. They include horizontal and vertical tangents without separate reciprocal formulas.
If both parameter derivatives vanish, the parametrization is stationary, and the velocity formula gives no direction. It does not prove that no tangent exists. The parametrization x = t³, y = t⁶ has zero velocity at t = 0, but its curve is y = x² and has the horizontal tangent y = 0. Eliminate the parameter or examine the first nonzero terms.
One geometric point can correspond to several parameter values. At a self-intersection, those parameter values may provide different tangent directions. A complete answer should identify the requested branch or list each applicable tangent rather than silently choosing one value of t.
ORIGINAL WORKED EXAMPLE
A parametric contact with a finite slope
For x = t² + 1 and y = t³ − t, find the tangent and normal at t = 2.
The point is (5,6). Differentiate with respect to t: x′ = 2t and y′ = 3t² − 1, giving direction (4,11).
The tangent slope is 11/4, so its equation is 4(y − 6) − 11(x − 5) = 0.
The normal is 4(x − 5) + 11(y − 6) = 0.
These simplify to 11x − 4y − 31 = 0 and 4x + 11y − 86 = 0. Both contain (5,6), and their direction vectors are perpendicular.
When a tangent must be parallel to a given nonvertical line of slope k, solve f′(a) = k and then compute the actual contact points and line equations. When a normal is parallel to that line, the tangent direction must be perpendicular to it. Use vectors if either line is vertical. Equality of slopes only determines direction, not coincidence.
For two nonvertical tangent slopes m₁ and m₂, the acute or zero angle φ between their unoriented lines satisfies tan φ = |(m₂ − m₁)/(1 + m₁m₂)| when the denominator is nonzero. If 1 + m₁m₂ = 0, the lines are perpendicular and φ = π/2. To avoid ambiguity, specify whether the problem requests the acute angle between lines or an oriented angle between moving directions.
The vector method works uniformly: for nonzero direction vectors u and v, cos φ = |u·v|/(|u||v|) gives the angle between unoriented lines in [0,π/2]. Without the absolute value it gives the angle between the chosen oriented vectors in [0,π]. At an intersection of two curves, first identify the correct common point, then evaluate each derivative at that point.
ORIGINAL WORKED EXAMPLE
Tangents with a specified direction
Find points on y = x³ − 3x whose tangents are parallel to y = 9x + 1, and give those tangents.
The required slope is 9. Differentiate the curve: y′ = 3x² − 3. Solve 3a² − 3 = 9 to obtain a = ±2.
The contacts are (2,2) and (−2,−2), since 8 − 6 = 2 and −8 + 6 = −2.
At (2,2), the tangent is y − 2 = 9(x − 2), giving y = 9x − 16.
At (−2,−2), the tangent is y + 2 = 9(x + 2), giving y = 9x + 16. Neither must share the given line’s intercept.
Answer Contacts (2,2), (−2,−2); tangents y = 9x − 16 and y = 9x + 16.
07 / UNDERSTAND & APPLY
7. Tangency, repeated intersections and local approximation
For a differentiable graph, subtract its tangent: R(x) = f(x) − f(a) − f′(a)(x − a). The remainder satisfies R(a) = 0 and R′(a) = 0. This expresses first-order agreement. If f is sufficiently smooth, higher derivatives determine whether the graph lies locally above, below or on opposite sides of the tangent. A positive second derivative at the contact places the curve locally above it; a cubic leading difference changes sign.
For a polynomial graph intersected with a line, tangency at a finite-slope contact implies that the difference polynomial has a repeated root. This gives an algebraic method: impose both P(a) = 0 and P′(a) = 0 for the intersection polynomial. A zero discriminant works directly when the resulting polynomial is quadratic, but do not apply a quadratic discriminant to a cubic or higher-degree equation.
The linear approximation f(a + h) ≈ f(a) + f′(a)h is a separate useful consequence. It becomes accurate relative to |h| as h tends to zero. The tangent is generally not an exact replacement for the whole curve, and the size of the error depends on curvature and distance from the contact. A tangent-line estimate far from the contact has no automatic accuracy guarantee.
A corner, a cusp and a vertical tangent are different phenomena. At a corner such as |x|, the one-sided slopes disagree. At a cusp such as y = |x|^(2/3), the slopes become unbounded in opposite signs but the limiting tangent line is vertical. At y = x^(1/3), both sides have positive unbounded slope and again a vertical tangent. Check line directions, not only the existence of a finite derivative.
08 / UNDERSTAND & APPLY
8. Original practice with contact checks
In each answer verify that the line contains the contact. When a vertical or horizontal line appears, write its equation directly instead of expressing it as an impossible slope.
ORIGINAL PRACTICE · 1
Find the tangent and normal to y = x³ at the origin.
Reveal answer & reasoning
Tangent y = 0; normal x = 0.
At x = 0, y = 0 and f′(0) = 0. A horizontal tangent has a vertical normal. The tangent crosses the cubic, which does not invalidate tangency.
ORIGINAL PRACTICE · 2
Find the tangent and normal to y = √x at x = 4.
Reveal answer & reasoning
Tangent y − 2 = (x − 4)/4; normal y − 2 = −4(x − 4).
The derivative 1/(2√x) is 1/4 at the contact (4,2). Its negative reciprocal is −4. Both formulas apply because x = 4 is in the positive differentiability domain.
ORIGINAL PRACTICE · 3
Find the tangent and normal to x² + y² = 25 at (5,0).
Reveal answer & reasoning
Tangent x = 5; normal y = 0.
The gradient is (10,0). The tangent is 10(x − 5) = 0 and the normal runs in the horizontal gradient direction. A slope quotient would divide by zero unnecessarily.
ORIGINAL PRACTICE · 4
Find the tangent and normal to x = cos t, y = 2 sin t at t = π/2.
Reveal answer & reasoning
Tangent y = 2; normal x = 0.
The contact is (0,2) and the velocity is (−1,0), giving a horizontal tangent and a vertical normal.
ORIGINAL PRACTICE · 5
Find the tangent to y = ln x parallel to y = 2x.
Reveal answer & reasoning
y = 2x − 1 − ln 2.
The derivative is 1/x on x > 0. Set 1/a = 2 to get a = 1/2, with f(a) = −ln 2. Substituting into point-slope form gives y + ln 2 = 2(x − 1/2).
ORIGINAL PRACTICE · 6
Find tangents to y = x² through (0,1).
Reveal answer & reasoning
There are no real tangents through (0,1).
The general tangent is y = 2ax − a². At x = 0 its intercept is −a² ≤ 0, so it cannot pass through a point with intercept 1.
ORIGINAL PRACTICE · 7
Find the acute angle between the tangents to y = x² and y = −x² + 2 at their intersection with positive x.
Reveal answer & reasoning
arctan(4/3).
Their positive-x intersection is (1,1). The slopes are 2 and −2. Therefore tan φ = |(−2 − 2)/(1 − 4)| = 4/3. The absolute value selects the acute angle.
ORIGINAL PRACTICE · 8
For x = t³, y = t⁶ at t = 0, does zero velocity imply that the tangent does not exist?
Reveal answer & reasoning
No; the tangent is y = 0.
Eliminating the parameter gives y = x² for all real x. That regular graph has derivative zero at the origin. The chosen parametrization pauses there, so its zero velocity alone cannot determine the direction.
09 / UNDERSTAND & APPLY
9. A compact checklist for every line
Write the contact, the direction and the line equation as three distinct steps. This prevents using a supplied external point as the contact or forgetting to evaluate the derivative at the correct parameter. Substitute the contact into your final equation; this catches many sign and intercept errors instantly.
Choose the representation that avoids artificial exceptions. Point-slope form is ideal for a finite derivative. The implicit gradient form handles vertical tangents. A parametric direction vector handles arbitrary regular parametrizations. At singular or stationary points, return to limits or the actual local branches rather than dividing zero by zero.
When asked for all tangents or normals satisfying a condition, solve for every allowed contact and deduplicate coincident lines. Check domain restrictions before accepting roots introduced by algebra. If the problem concerns angles, state the angle convention. If it concerns an approximation, state that the tangent is local and check how close the evaluation point is.
The normal passes through the same contact as the tangent. It is not generally the radius from the origin, except for special curves such as circles centered there. This distinction matters for ellipses, parabolas and shifted circles. Perpendicularity is a direction condition, not a license to choose an arbitrary line with the correct slope.
KEEP THE BIG IDEAS
Quick revision
A tangent is local direction and may cross the curve.
Use a point plus a direction; verify the contact lies in the original domain.
Implicit gradients and parametric vectors avoid vertical-line division errors.
Zero gradient or zero parameter velocity requires further analysis, not an automatic conclusion.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.