Tangent and normal to a curve — how to find equations step by step

Question

Find the equations of the tangent and normal to the curve $y = x^3 - 3x + 2$ at the point where $x = 1$ .

(CBSE 12 + JEE Main pattern)

Solution — Step by Step

Substitute $x = 1$ into $y = x^3 - 3x + 2$ :

y = 1 - 3 + 2 = 0

The point is $(1, 0)$ .

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

At $x = 1$ : slope of tangent $= 3(1)^2 - 3 = 0$ .

The tangent is horizontal at this point.

Using point-slope form $y - y_1 = m(x - x_1)$ :

y - 0 = 0(x - 1)

\mathbf{y = 0}

The tangent is the x-axis itself.

The normal is perpendicular to the tangent. Since the tangent is horizontal (slope = 0), the normal is vertical:

\mathbf{x = 1}

When the tangent slope is 0, the normal slope is undefined (vertical line). When the tangent slope is $m$ , the normal slope is $-\dfrac{1}{m}$ .

flowchart TD
    A["Given: curve and point"] --> B["Step 1: Find y-coordinate"]
    B --> C["Step 2: Compute dy/dx"]
    C --> D["Step 3: Evaluate dy/dx at the point = slope m"]
    D --> E["Tangent: y - y₁ = m(x - x₁)"]
    D --> F["Normal slope = -1/m"]
    F --> G["Normal: y - y₁ = (-1/m)(x - x₁)"]
    E --> H["Special case: m = 0 → tangent is horizontal"]
    F --> I["Special case: m = 0 → normal is vertical"]

Why This Works

The derivative $\dfrac{dy}{dx}$ at a point gives the slope of the tangent line at that point. This is the fundamental geometric meaning of the derivative — it measures how steeply the curve is rising or falling at any location.

The normal is perpendicular to the tangent. Two perpendicular lines have slopes that multiply to $-1$ : if tangent slope is $m$ , normal slope is $-1/m$ . Once we have the slope and a point, the point-slope form gives the complete equation.

Alternative Method — Implicit Differentiation for Implicit Curves

If the curve is given implicitly (like $x^2 + y^2 = 25$ ), differentiate both sides with respect to $x$ :

2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}

Then proceed as before. This is essential for JEE problems involving circles, ellipses, and other implicit curves.

For JEE Main, tangent-normal problems often involve finding the tangent at a given slope (not a given point). If the question says “find the tangent to $y = x^2$ with slope 6,” set $\dfrac{dy}{dx} = 6$ , which gives $2x = 6 \implies x = 3$ , then $y = 9$ . Tangent: $y - 9 = 6(x - 3)$ , or $y = 6x - 9$ .

Common Mistake

Students forget to find the y-coordinate. They compute the slope correctly but then write the tangent equation using only the x-value without substituting back to find $y$ . The point-slope form needs BOTH coordinates $(x_1, y_1)$ . Also, when the tangent slope is 0 (horizontal tangent), the normal is vertical ( $x = x_1$ ), not " $y = -\frac{1}{0}$ " — division by zero means a vertical line, not infinity.

Question

Solution — Step by Step

Why This Works

Alternative Method — Implicit Differentiation for Implicit Curves

Common Mistake

Try These Next