Find the equation of normal to curve y = x³ - 3x at point (1, -2)

Question

Find the equation of the normal to the curve $y = x^3 - 3x$ at the point $(1, -2)$ .

(NCERT Class 12, Chapter 6 — Application of Derivatives)

Solution — Step by Step

At $x = 1$ : $y = 1^3 - 3(1) = 1 - 3 = -2$ ✓

The point $(1, -2)$ is indeed on the curve.

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

At $x = 1$ : $\frac{dy}{dx} = 3(1)^2 - 3 = 3 - 3 = 0$

The slope of the tangent at $(1, -2)$ is $0$ (horizontal tangent).

The normal is perpendicular to the tangent. If the tangent slope is $m_t$ , then the normal slope is $m_n = -1/m_t$ .

But here $m_t = 0$ (horizontal tangent), so the normal is vertical. A vertical line has an undefined slope.

A vertical line passing through $(1, -2)$ has the equation:

\boxed{x = 1}

Why This Works

The tangent at $(1, -2)$ is horizontal because the derivative is zero there — this means the curve has a local maximum or minimum at this point. (In fact, $x = 1$ is a local minimum of $y = x^3 - 3x$ .)

Since the normal is always perpendicular to the tangent, a horizontal tangent produces a vertical normal. The equation of a vertical line is simply $x = \text{constant}$ , where the constant is the x-coordinate of the given point.

For the general case (non-zero tangent slope $m_t$ ), the normal equation would be:

y - y_1 = -\frac{1}{m_t}(x - x_1)

Alternative Method — Direct approach for any point

If you want the normal at a general point $(a, a^3 - 3a)$ :

Tangent slope: $m_t = 3a^2 - 3$

Normal slope: $m_n = -\frac{1}{3a^2 - 3}$ (provided $a \neq \pm 1$ )

Normal equation: $y - (a^3 - 3a) = -\frac{1}{3a^2 - 3}(x - a)$

When the tangent slope comes out as $0$ , many students get stuck because $-1/0$ is undefined. Don’t panic — a zero tangent slope simply means a vertical normal. Write $x = x_1$ and you’re done. Similarly, if the tangent is vertical (slope undefined), the normal is horizontal: $y = y_1$ .

Common Mistake

The most common error: students compute the tangent slope as $0$ and then write the normal equation as $y = -2$ (confusing tangent and normal). The tangent is $y = -2$ (horizontal line). The normal, being perpendicular to it, is $x = 1$ (vertical line). Always double-check: tangent and normal must be perpendicular to each other.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct approach for any point

Common Mistake

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