Tangent and Normal to a Curve — How to Find Equations Step by Step

Question

Given a curve $y = f(x)$ and a point on it, how do we find the equations of the tangent and the normal?

Solution — Step by Step

The slope of the tangent at any point $(x_0, y_0)$ on the curve $y = f(x)$ is:

m_{\text{tangent}} = \left.\frac{dy}{dx}\right|_{x = x_0}

Differentiate $f(x)$ , then substitute $x = x_0$ to get the numerical slope.

Example: For $y = x^3 - 3x + 2$ at $x = 1$ :

\frac{dy}{dx} = 3x^2 - 3

m_{\text{tangent}} = 3(1)^2 - 3 = 0

The normal is perpendicular to the tangent. For perpendicular lines:

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

If $m_{\text{tangent}} = 0$ (horizontal tangent), the normal is vertical: $x = x_0$ .

If $m_{\text{tangent}}$ is undefined (vertical tangent), the normal is horizontal: $y = y_0$ .

With slope $m$ and point $(x_0, y_0)$ :

Tangent: $y - y_0 = m_{\text{tangent}}(x - x_0)$

Normal: $y - y_0 = m_{\text{normal}}(x - x_0)$

For our example: $y_0 = 1^3 - 3(1) + 2 = 0$ , $m_{\text{tangent}} = 0$ .

Tangent: $y - 0 = 0(x - 1) \implies y = 0$

Normal: $x = 1$ (vertical line, since tangent is horizontal).

graph TD
    A[Given curve y = f x and point x0] --> B[Find y0 = f of x0]
    B --> C[Differentiate: dy/dx = f prime x]
    C --> D[Evaluate m = f prime at x0]
    D --> E{Is m finite and nonzero?}
    E -->|Yes| F[Tangent: y - y0 = m times x - x0]
    E -->|m = 0| G[Tangent: y = y0, Normal: x = x0]
    E -->|m undefined| H[Tangent: x = x0, Normal: y = y0]
    F --> I[Normal: y - y0 = -1/m times x - x0]

Why This Works

The derivative $\frac{dy}{dx}$ at a point gives the instantaneous rate of change — which is geometrically the slope of the tangent line at that point. This is the fundamental connection between calculus and coordinate geometry.

The normal being perpendicular means their slopes multiply to $-1$ (the perpendicularity condition from coordinate geometry).

For CBSE 12 boards, tangent-normal problems carry 4-5 marks. The marking scheme gives marks for: finding the derivative (1 mark), evaluating at the point (1 mark), writing tangent equation (1 mark), and writing normal equation (1 mark). Show every step clearly.

Alternative Method

If the curve is given in implicit form $F(x, y) = 0$ , use implicit differentiation:

\frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}

For example, $x^2 + y^2 = 25$ at $(3, 4)$ :

\frac{dy}{dx} = -\frac{2x}{2y} = -\frac{3}{4}

Tangent: $y - 4 = -\frac{3}{4}(x - 3) \implies 3x + 4y = 25$ .

Common Mistake

Students sometimes find the slope at $x = x_0$ but forget to compute $y_0 = f(x_0)$ and use the wrong $y$ -coordinate. For example, if the question says “find the tangent at $x = 2$ ,” you must compute $y = f(2)$ first. Using the given $x$ but guessing $y$ (or using $y = 2$ ) gives the wrong tangent line.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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