MathsISC Class 12 · Applications of Derivatives

Tangents and Normals: Equations, Special Conditions and Angle Between Curves

Updated 15 October 2026 · Fact-checked

The slope of the tangent to y = f(x) at a point is dy/dx there, so the tangent at (x₁, y₁) is y − y₁ = m(x − x₁) with m = f′(x₁). The normal is perpendicular to it, with slope −1/m. A tangent is parallel to the x-axis where dy/dx = 0.

The tangent at P has slope m = dy/dx, and the normal is at right angles to it, so its slope is −1/m.

Illustration drawn for QuestionBuddy.

Understand Tangents and Normals

Slope of the tangent. The derivative $\frac{d y}{d x}$ at a point is the slope of the tangent to the curve at that point. This geometrical meaning is the key to the whole topic; the ISC 2024 and 2025 examiners both asked candidates to use it and noted that some could not link $\frac{d y}{d x}$ with the slope of a line.

Equations. At the point $P (x_{1}, y_{1})$ on $y = f (x)$ , let $m = \frac{d y}{d x}_{P}$ .

Tangent: $y - y_{1} = m (x - x_{1})$ .
Normal (perpendicular to the tangent at $P$ ): $y - y_{1} = - \frac{1}{m} (x - x_{1})$ , if $m \neq = 0$ .
If $m = 0$ , the tangent is horizontal ( $y = y_{1}$ ) and the normal is vertical ( $x = x_{1}$ ).
If $\frac{d y}{d x}$ is infinite (that is, $\frac{d x}{d y} = 0$ ), the tangent is vertical ( $x = x_{1}$ ) and the normal horizontal.

Special conditions. Many questions ask for the point where the tangent has a given direction:

parallel to the x-axis: $\frac{d y}{d x} = 0$ ;
parallel to the y-axis: $\frac{d x}{d y} = 0$ ;
parallel to a given line $a x + b y + c = 0$ : $\frac{d y}{d x} = - \frac{a}{b}$ ;
perpendicular to that line: $\frac{d y}{d x} = \frac{b}{a}$ ;
parallel to the chord joining two points of the curve: $\frac{d y}{d x}$ = slope of the chord (ISC 2024 asked this for $y = (x - 2)^{2}$ ).

Solve for $x$ , then find $y$ from the equation of the curve, not from the derivative.

Implicit and parametric curves. For a relation such as $\frac{x ^{2}}{9} + \frac{y ^{2}}{16} = 1$ , find $\frac{d y}{d x}$ implicitly. For $x = f (t)$ , $y = g (t)$ , use $\frac{d y}{d x} = \frac{g ^{'} ( t )}{f ^{'} ( t )}$ and the point $(f (t), g (t))$ . For the parabola $x = a t^{2}$ , $y = 2 a t$ the tangent at parameter $t$ simplifies to $t y = x + a t^{2}$ .

Angle between two curves. Two curves meeting at a point make the angle between their tangents there. If the slopes are $m_{1}$ and $m_{2}$ ,
$tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} .$
The curves cut orthogonally (at right angles) when $m_{1} m_{2} = - 1$ , and touch each other when $m_{1} = m_{2}$ . To find the angle, first find the point of intersection by solving the two equations together, then the two slopes there.

Why the normal matters. In physics and engineering, forces of contact act along the normal, and a light ray striking a curved mirror along the normal is reflected straight back. In geometry, the normal to a circle at any point passes through its centre.

Tangents and Normals: formulas to remember

Slope of tangent: $m = \frac{d y}{d x}_{(x_{1}, y_{1})}$; Evaluate the derivative at the point.
Tangent: $y - y_{1} = m (x - x_{1})$; Point-slope form through the point of contact.
Normal: $y - y_{1} = - \frac{1}{m} (x - x_{1})$; m ≠ 0; if m = 0 the normal is x = x₁.
Horizontal and vertical tangents: $\frac{d y}{d x} = 0 (parallel to x-axis), \frac{d x}{d y} = 0 (parallel to y-axis)$; Find y from the curve, not from the derivative.
Parallel to ax + by + c = 0: $\frac{d y}{d x} = - \frac{a}{b}$; Perpendicular: dy/dx = b/a.
Angle between curves: $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$; Orthogonal if m₁m₂ = −1.
Parabola tangent: $x = a t^{2}, y = 2 a t : t y = x + a t^{2}$; Slope 1/t.

How to solve Tangents and Normals questions

Most tangent and normal questions follow this routine.

1Find $\frac{d y}{d x}$ for the curve (directly, implicitly or parametrically).
2If the point is given, substitute it to get the slope $m$ . If a condition is given instead (parallel to a line, horizontal, parallel to a chord), set $\frac{d y}{d x}$ equal to the required slope and solve for $x$ .
3Find the corresponding $y$ from the equation of the curve, and check the point lies on it.
4Write the tangent $y - y_{1} = m (x - x_{1})$ and simplify to the form $a x + b y + c = 0$ .
5For the normal, use slope $- \frac{1}{m}$ (or a vertical or horizontal line if $m = 0$ or undefined).
6For the angle between two curves, solve their equations together for the intersection point, find both slopes there, and use $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$ .
7Give the final equations neatly; examiners mark the final form as well as the slope.

Tangents and Normals in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Tangent line as a direction of motion

Suppose a car drives along a curved road whose shape on a map is $y = \frac{x ^{2}}{100}$ (in metres). Its headlights shine straight ahead, along the tangent to the road.

When the car is at $x = 50$ , find the line along which the headlights shine. Will the beam light up a sign at $(80, 55)$ ?

Try it first, then show the solutionSolution

$\frac{d y}{d x} = \frac{x}{50}$ ; at $x = 50$ , slope $= 1$ and $y = 25$ .
Beam (tangent): $y - 25 = 1 (x - 50)$ , so $y = x - 25$ .
At $x = 80$ : $y = 55$ . The sign lies on the beam.

Answer: The beam follows $y = x - 25$ , and it does light up the sign at $(80, 55)$ .

Memory hook: Headlights point along the tangent: slope = dy/dx at the car.

Normal to a curve

Imagine a curved reflector whose cross-section is the parabola $y^{2} = 4 x$ (in decimetres). A ray of light that strikes the reflector along the normal is reflected straight back along the same line.

Find the equation of the line along which a ray must strike the point $(1, 2)$ to be reflected straight back.

Try it first, then show the solutionSolution

Differentiate implicitly: $2 y \frac{d y}{d x} = 4$ , so $\frac{d y}{d x} = \frac{2}{y} = 1$ at $(1, 2)$ .
Normal slope $= - 1$ . Normal: $y - 2 = - (x - 1)$ , so $x + y = 3$ .

Answer: Along the line $x + y = 3$ .

Memory hook: Straight back means along the normal: slope −1/m.

Tricks to remember Tangents and Normals

Short memory aids. Say each one aloud once, then use it on a question the same day.

Flip and change sign
Normal slope = negative reciprocal of tangent slope: 2 becomes −1/2.
Use it when: Writing the normal quickly.
Curve for y, derivative for slope
Use the derivative only for the slope; always get y from the curve itself.
Use it when: Avoiding wrong points in "parallel to the x-axis" questions.
Flat means zero
Parallel to the x-axis: dy/dx = 0. Parallel to the y-axis: dx/dy = 0.
Use it when: Special-direction questions.
Orthogonal = product −1
Curves cut at right angles when the product of their slopes at the meeting point is −1.
Use it when: Proofs of orthogonal intersection.

Tangents and Normals: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Slope of a normal · 1 mark · MCQBoard level

The slope of the normal to $y = x^{2} + 3 x$ at $x = 1$ is:
(a) 5
(b) $- 5$
(c) $\frac{1}{5}$
(d) $- \frac{1}{5}$

Try it first, then show the solutionSolution

Tangent slope $= 2 x + 3 = 5$ ; normal slope $= - \frac{1}{5}$ .

Answer: (d) $- \frac{1}{5}$ .

Example 2: Horizontal tangents to a cubic · 2 marks · Short answerBoard level

Find the points on the curve $y = x^{3} - 3 x^{2} - 9 x + 7$ at which the tangent is parallel to the x-axis.

Try it first, then show the solutionSolution

$\frac{d y}{d x} = 3 x^{2} - 6 x - 9 = 3 (x - 3) (x + 1) = 0 \Rightarrow x = 3$ or $x = - 1$ . (1 mark)
From the curve: at $x = 3$ , $y = 27 - 27 - 27 + 7 = - 20$ ; at $x = - 1$ , $y = - 1 - 3 + 9 + 7 = 12$ . The points are $(3, - 20)$ and $(- 1, 12)$ . (1 mark)

Answer: $(3, - 20)$ and $(- 1, 12)$ . (ISC 2025 asked the same type for $y = 2 x^{2} - 6 x - 4$ .)

Example 3: Tangent and normal at a point · 3 marks · Short answerHarder board level

Find the equations of the tangent and normal to $y = x^{3} - 2 x + 3$ at the point $(1, 2)$ .

Try it first, then show the solutionSolution

$\frac{d y}{d x} = 3 x^{2} - 2 = 1$ at $x = 1$ . (1 mark)
Tangent: $y - 2 = 1 (x - 1) \Rightarrow x - y + 1 = 0$ . (1 mark)
Normal: $y - 2 = - 1 (x - 1) \Rightarrow x + y - 3 = 0$ . (1 mark)

Answer: Tangent $x - y + 1 = 0$ ; normal $x + y - 3 = 0$ .

Example 4: Tangent parallel to a given line · 3 marks · Short answerHarder board level

Find the equation of the tangent to $y = x^{2} - 2 x + 7$ that is parallel to the line $2 x - y + 5 = 0$ .

Try it first, then show the solutionSolution

The line has slope 2. Set $\frac{d y}{d x} = 2 x - 2 = 2 \Rightarrow x = 2$ . (1 mark)
From the curve, $y = 4 - 4 + 7 = 7$ . (1 mark)
Tangent: $y - 7 = 2 (x - 2) \Rightarrow 2 x - y + 3 = 0$ . (1 mark)

Answer: $2 x - y + 3 = 0$ .

Example 5: Angle between two parabolas · 3 marks · Short answerHarder board level

Find the angle between the curves $y = x^{2}$ and $x = y^{2}$ at the point $(1, 1)$ .

Try it first, then show the solutionSolution

For $y = x^{2}$ : $m_{1} = 2 x = 2$ at $(1, 1)$ . (1 mark)
For $x = y^{2}$ : $1 = 2 y \frac{d y}{d x}$ , so $m_{2} = \frac{1}{2 y} = \frac{1}{2}$ at $(1, 1)$ . (1 mark)
$tan θ = \frac{2 - \frac{1}{2}}{1 + 1} = \frac{3}{4}$ , so $θ = tan^{- 1} \frac{3}{4}$ . (1 mark)

Answer: $θ = tan^{- 1} \frac{3}{4}$ (about $36. 9^{\circ}$ ).

Example 6: Condition for orthogonal curves · 5 marks · Long answerHOTS / competency

Prove that the curves $x = y^{2}$ and $x y = k$ cut at right angles if $8 k^{2} = 1$ .

Try it first, then show the solutionSolution

Intersection: substitute $x = y^{2}$ in $x y = k$ : $y^{3} = k$ , so $y = k^{1/3}$ and $x = k^{2/3}$ . (1 mark)
For $x = y^{2}$ : $m_{1} = \frac{1}{2 y}$ . For $x y = k$ : $y + x \frac{d y}{d x} = 0$ , so $m_{2} = - \frac{y}{x}$ . (2 marks)
Orthogonal if $m_{1} m_{2} = - 1$ : $\frac{1}{2 y} \cdot (- \frac{y}{x}) = - \frac{1}{2 x} = - 1$ , so $x = \frac{1}{2}$ . (1 mark)
Then $k^{2/3} = \frac{1}{2} \Rightarrow k^{2} = \frac{1}{8} \Rightarrow 8 k^{2} = 1$ . (1 mark)

Answer: Proved: the curves are orthogonal exactly when $8 k^{2} = 1$ .

Where students lose marks in Tangents and Normals

The slips examiners see most often, and what to write instead.

Mistake: Not equating dy/dx to the required slope
The ISC 2024 and 2025 examiners reported candidates who did not equate the derivative to the slope of the chord, or to zero.
Do this: Write the condition in words first ("tangent parallel to x-axis means dy/dx = 0") and then in symbols.
Usually costs: The whole 2-mark question.
Mistake: Finding y from the derivative
Substituting x into dy/dx gives the slope, not the y-coordinate of the point.
Wrong working
$x = 3$ : $y = 3 x^{2} - 6 x - 9 = 0$ , so $(3, 0)$ .
Right working
$x = 3$ : $y = 3^{3} - 3 (3)^{2} - 9 (3) + 7 = - 20$ , so $(3, - 20)$ .
Do this: Get y from the equation of the curve.
Usually costs: 1 mark.
Mistake: Using m instead of −1/m for the normal
The normal is perpendicular to the tangent, not parallel to it.
Do this: Write "slope of normal = −1/m" as a separate line.
Usually costs: 1 mark.
Mistake: Computational slips in finding the point
The 2025 examiners noted computational errors after correctly setting dy/dx = 0.
Do this: Check the point satisfies the curve equation before writing the line.
Usually costs: 1 mark.

Exam tips

State the slope as a separate line ("slope of tangent at $(1, 2)$ is 1"); it usually earns a mark even if the line is miswritten.
Present final equations in the form $a x + b y + c = 0$ with integer coefficients.
In "parallel to a chord" questions, find the chord slope first from the two given points.
For angle-between-curves questions, give the angle as $tan^{- 1} (\dots)$ unless a standard angle results.
If $\frac{d y}{d x}$ is undefined at the point, the tangent is vertical: write $x = x_{1}$ .

Practice questions on Tangents and Normals

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The equation of the tangent to $y = e^{x}$ at $x = 0$ is:
(a) $y = x$
(b) $y = x + 1$
(c) $y = 1$
(d) $y = e x$
Hint
Slope $e^{0} = 1$ at $(0, 1)$ .
Answer
(b) $y = x + 1$ .
Q2 · 2 marks · Short answerBoard level
Find the equation of the normal to $y = 2 x^{2}$ at $(1, 2)$ .
Hint
Tangent slope 4.
Answer
Normal slope $- \frac{1}{4}$ : $y - 2 = - \frac{1}{4} (x - 1)$ , so $x + 4 y - 9 = 0$ .
Q3 · 3 marks · Short answerHarder board level
Find the points on the ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{16} = 1$ where the tangent is (i) parallel to the x-axis, (ii) parallel to the y-axis.
Hint
$\frac{d y}{d x} = - \frac{16 x}{9 y}$ .
Answer
(i) $\frac{d y}{d x} = 0 \Rightarrow x = 0$ , so $(0, \pm 4)$ . (ii) $\frac{d x}{d y} = - \frac{9 y}{16 x} = 0 \Rightarrow y = 0$ , so $(\pm 3, 0)$ .
Q4 · 2 marks · Short answerHarder board level
Find the point on $y = x^{2}$ at which the tangent is parallel to the chord joining $(1, 1)$ and $(3, 9)$ .
Hint
Chord slope $= \frac{9 - 1}{3 - 1}$ .
Answer
Chord slope 4; $2 x = 4 \Rightarrow x = 2$ , so the point is $(2, 4)$ .
Q5 · 3 marks · Short answerHarder board level
Find the equation of the tangent to the curve $x = a t^{2}$ , $y = 2 a t$ at the point with parameter $t$ , and hence at $t = 2$ when $a = 1$ .
Hint
$\frac{d y}{d x} = \frac{1}{t}$ .
Answer
$y - 2 a t = \frac{1}{t} (x - a t^{2}) \Rightarrow t y = x + a t^{2}$ . With $a = 1$ , $t = 2$ : $2 y = x + 4$ , at the point $(4, 4)$ .
Q6 · 3 marks · Short answerHOTS / competency
Find the equation of the tangent to $y = 3 x - 2$ that is parallel to the line $4 x - 2 y + 5 = 0$ .
Hint
Slope 2; solve $\frac{3}{2 3 x - 2} = 2$ .
Answer
$3 x - 2 = \frac{3}{4} \Rightarrow x = \frac{41}{48}$ , $y = \frac{3}{4}$ . Tangent: $y - \frac{3}{4} = 2 (x - \frac{41}{48})$ , that is $48 x - 24 y = 23$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the side profile of a skateboard ramp is $y = 0.25 x^{2}$ for $0 \leq x \leq 4$ (in metres).
(i) Find the slope of the ramp at $x = 2$ and the angle it makes with the ground there. (1 mark)
(ii) Find the equation of the tangent at $x = 2$ . (1 mark)
(iii) Find the equation of the normal at $x = 2$ . (1 mark)
(iv) At which point is the slope $\frac{1}{2}$ ? (1 mark)
Hint
$\frac{d y}{d x} = 0.5 x$ .
Answer
(i) Slope 1, angle $4 5^{\circ}$ . (ii) At $(2, 1)$ : $y = x - 1$ . (iii) $y - 1 = - (x - 2)$ , so $x + y = 3$ . (iv) $0.5 x = \frac{1}{2} \Rightarrow x = 1$ : the point $(1, 0.25)$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does $\frac{d y}{d x}$ at a point represent geometrically?Show answer
The slope of the tangent to the curve at that point.
2.Write the equation of the normal at $(x_{1}, y_{1})$ with tangent slope $m$ .Show answer
$y - y_{1} = - \frac{1}{m} (x - x_{1})$ .
3.When is a tangent parallel to the y-axis?Show answer
When $\frac{d x}{d y} = 0$ .
4.State the condition for two curves to cut orthogonally.Show answer
The product of their slopes at the point of intersection is $- 1$ .
5.Find the slope of the tangent to $y = sin x$ at $x = 0$ .Show answer
1.
6.Write the tangent to $x = a t^{2}$ , $y = 2 a t$ at parameter $t$ .Show answer
$t y = x + a t^{2}$ .

Tangents and Normals: frequently asked questions

How do I find the equation of a tangent to a curve?

Differentiate to get dy/dx, substitute the point to get the slope m, then use the point-slope form y − y₁ = m(x − x₁).

What is the difference between a tangent and a normal?

The tangent touches the curve at the point and has slope dy/dx there. The normal passes through the same point at right angles to the tangent, so its slope is the negative reciprocal.

How do I find the angle between two curves?

Find their point of intersection, find the slope of each curve there, and use tan θ = |(m₁ − m₂)/(1 + m₁m₂)|. If m₁m₂ = −1, they cut at right angles.

What if the slope of the tangent is zero?

Then the tangent is horizontal, y = y₁, and the normal is the vertical line x = x₁. You cannot use −1/m for the normal in that case.

Is the angle between two curves in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus lists the equation of tangent and normal and the angle between two curves under Applications of Derivatives.