On this page
Tangent and normal questions are short, but each has a trap: the normal slope with the wrong sign, a vertical tangent hiding in a parametric curve, or a "tangent from a point" treated as if that point were on the curve. This guide works through each type with checked numbers and finishes with a practice plan.
Which exams include tangents and normals?
The JEE (Advanced) 2026 brochure lists "tangents and normals" under differential calculus, and tangents and normals to circles and conics under coordinate geometry. The JEE (Main) 2026 syllabus lists applications of derivatives as rate of change, increasing and decreasing functions, and maxima and minima, without naming tangents and normals. The CBSE Mathematics 2026-27 curriculum lists the same three applications for Class XII. Check your year's documents, since these lists change.
How do you write the tangent and normal?
At a point (x1, y1) on y = f(x):
Tangent: y − y1 = m(x − x1), with m = f′(x1).
Normal: y − y1 = −(1/m)(x − x1).
If m = 0, the tangent is horizontal (y = y1) and the normal vertical (x = x1).
Example 1. Find the tangent and normal to y = x³ − 3x + 2 at x = 2.
y(2) = 8 − 6 + 2 = 4. The point is (2, 4).
dy/dx = 3x² − 3, so m = 3(4) − 3 = 9.
Tangent: y − 4 = 9(x − 2), so y = 9x − 14.
Normal: y − 4 = −(1/9)(x − 2). Multiply by 9: 9y − 36 = −x + 2, so x + 9y − 38 = 0.
Check: (2, 4) is on both lines. 9(2) − 14 = 4, and 2 + 36 − 38 = 0.
Where is the tangent horizontal? Set 3x² − 3 = 0, so x = ±1. The points are (1, 0) and (−1, 4).
Does a tangent touch the curve at only one point?
Not always, and this is a real misconception. Biza, Christou and Zachariades (2008) surveyed 196 Year 12 students and described a shift from a "geometrical global" view of tangents, learned from circles, to an "analytical local" view, in which the tangent only has to match the curve's slope at one point.
Example 1 shows why the local view is needed. Solve x³ − 3x + 2 = 9x − 14: x³ − 12x + 16 = 0, which factors as (x − 2)²(x + 4) = 0. The tangent at x = 2 meets the curve again at x = −4. It is still the tangent at (2, 4).
How do you handle parametric curves?
Use dy/dx = (dy/dt) ÷ (dx/dt).
Example 2. x = t², y = 2t (the parabola y² = 4x). Find the tangent and normal at t = 2.
Point: (4, 4).
dy/dx = 2 ÷ (2t) = 1/t = 1/2 at t = 2.
Tangent: y − 4 = (1/2)(x − 4), so x − 2y + 4 = 0.
Normal slope −2: y − 4 = −2(x − 4), so 2x + y − 12 = 0.
Check with the standard result yy1 = 2(x + x1) for y² = 4x: 4y = 2(x + 4), which is x − 2y + 4 = 0 again. At t = 0, dx/dt = 0, so the tangent is vertical: the y-axis.
Tangents from a point outside the curve
Example 3. Find the tangents to y = x² that pass through (1, −3).
(1, −3) is not on the curve, since 1² ≠ −3.
Let the point of contact be (a, a²). The tangent there is y − a² = 2a(x − a).
It passes through (1, −3): −3 − a² = 2a − 2a², so a² − 2a − 3 = 0, giving (a − 3)(a + 1) = 0.
a = 3: y = 6x − 9. a = −1: y = −2x − 1.
Check: both pass through (1, −3), since 6 − 9 = −3 and −2 − 1 = −3.
How do you find the angle between two curves?
Find where they meet, then the angle between their tangents there: tan θ = |(m1 − m2) ÷ (1 + m1m2)|.
Example 4. y = x² and y = √x meet at (0, 0) and (1, 1).
At (1, 1): slopes are 2 and 1/2. tan θ = |(2 − 0.5) ÷ (1 + 1)| = 0.75, so θ = tan⁻¹(3/4), about 36.9°.
At (0, 0): y = x² has slope 0 (horizontal), and y = √x has a vertical tangent. The curves cut at 90°. The formula fails here because one slope is undefined, so reason from the directions.
What are the common slips?
Normal slope sign. It is −1/m, not 1/m or −m.
Using a point not on the curve as the point of contact.
Missing vertical tangents where dx/dt = 0 or the derivative is undefined.
Slope at the wrong point. Substitute the point of contact, not the outside point.
Leaving the answer in slope form when the question asks for a line equation. Write it fully.
A practice routine
Day 1: Six tangent-and-normal pairs, each checked by substituting the point.
Day 2: Four parametric curves, including one with a vertical tangent.
Day 3: Three tangents from outside points, and two angle-between-curves questions.
Day 5: A mixed set from previous JEE Advanced papers.
Tangents connect to the rest of the derivatives chapter: see our guide to maxima and minima and the mean value theorems. If you want to watch a tangent line move along a curve, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
Today, find the tangent to y = x³ − 3x + 2 at x = 0 and check where else it meets the curve.
