Tangents and Normals

Two related slopes

At a point on a curve $y=f(x)$ , the derivative $f'(x)$ gives the slope of the tangent line, the line that just touches the curve there. The normal is perpendicular to the tangent at that same point, so its slope is the tangent slope's negative reciprocal:

m_{\text{normal}} = -\dfrac{1}{m_{\text{tangent}}}

Equations of the tangent and normal

Once you know the slope $m$ at a point $(x_1, y_1)$ on the curve, the line through that point is $y - y_1 = m(x - x_1)$ . So at $x = x_1$ :

\text{Tangent: } y - y_1 = f'(x_1)(x - x_1)

\text{Normal: } y - y_1 = -\dfrac{1}{f'(x_1)}(x - x_1) \quad (f'(x_1) \ne 0)

How the tangent and normal behave in the three possible cases

At the point	Tangent	Normal
$f'(x_1) = m \ne 0$	slope $m$	slope $-\dfrac{1}{m}$
$f'(x_1) = 0$	horizontal: $y = y_1$	vertical: $x = x_1$
$\dfrac{dy}{dx} \to \infty$ (vertical tangent)	vertical: $x = x_1$	horizontal: $y = y_1$

Worked example: a cubic curve

Find the slope of the tangent to $y = x^3 - x$ at $x=2$ .

Solution: $y' = 3x^2 - 1$ . At $x=2$ :

y'(2) = 3(4) - 1 = 11

Worked example: a rational function

Find the slope of the tangent to $y = \dfrac{x-1}{x-2}$ (for $x \ne 2$ ) at $x=10$ .

Solution: By the quotient rule:

y' = \dfrac{(1)(x-2) - (x-1)(1)}{(x-2)^2} = \dfrac{-1}{(x-2)^2}

At $x=10$ :

y'(10) = \dfrac{-1}{(10-2)^2} = -\dfrac{1}{64}

Worked example: the slope of a normal

Find the slope of the normal to the curve $y = 2x^2 - 3\sin x$ at $x=0$ .

Solution: The tangent slope is the derivative:

y' = 4x - 3\cos x

At $x=0$ : $y'(0) = 0 - 3(1) = -3$ . The normal slope is the negative reciprocal:

m_{\text{normal}} = -\dfrac{1}{-3} = \dfrac{1}{3}

Worked example: both equations at a point

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

Solution: $\dfrac{dy}{dx} = 2x$ , so the tangent slope at $x = 1$ is $2$ . The tangent is:

y - 1 = 2(x - 1) \;\Rightarrow\; y = 2x - 1

The normal slope is $-\dfrac{1}{2}$ , so the normal is:

y - 1 = -\dfrac{1}{2}(x - 1) \;\Rightarrow\; x + 2y = 3

y = x² with its tangent y = 2x − 1 and normal x + 2y = 3 at (1, 1). The two lines meet the curve at the same point and cross at a right angle.

Worked example: a tangent parallel to a line

Find the point on $y = x^2$ where the tangent is parallel to $y = 4x + 1$ , and the tangent's equation.

Solution: Parallel lines have equal slopes, so $\dfrac{dy}{dx} = 2x = 4$ , giving $x = 2$ and the point $(2, 4)$ . The tangent is $y - 4 = 4(x - 2)$ , that is $y = 4x - 4$ .

Common mistakes

Using $f'(x)$ instead of $f'(x_1)$ . The slope in the line's equation must be a number, the derivative evaluated at the point, not the general derivative.
Dropping the minus sign. The normal slope is the negative reciprocal. If the tangent slope is $-3$ , the normal slope is $+\dfrac{1}{3}$ .
Dividing by zero when the tangent is horizontal. If $f'(x_1) = 0$ , don't write $-\dfrac{1}{0}$ . The normal is simply the vertical line $x = x_1$ .
Using the wrong point. If you are only given $x_1$ , find $y_1 = f(x_1)$ from the curve's equation before writing either line.

Where this comes up in exams

JEE Advanced 2026 lists tangents and normals among the applications of derivatives. The 2025-26 CBSE Class XI syllabus relates the derivative to the slope of the tangent to a curve; the Class XII applications of derivatives (rate of change, increasing and decreasing functions, maxima and minima) and the JEE Main 2026 syllabus don't name tangents and normals.

Checked against JEE (Advanced) 2026 syllabus, CBSE Mathematics curriculum, Classes XI-XII (2025-26) and JEE (Main) 2026 syllabus (NTA). Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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