WAEC 2023 · Paper 1 · Q28

Applications of differentiation

Find the equation of the normal to the curve $y = 3x^2 + 2$ at point $(1, 5)$ .

Worked solution (try it first)

The gradient of the tangent is $\dfrac{dy}{dx} = 6x$ , which is 6 at $x = 1$ .
The normal is perpendicular to the tangent, so its gradient is $-\frac{1}{6}$ .
Through $(1, 5)$ : $y - 5 = -\frac{1}{6}(x - 1)$ , so $6y - 30 = -x + 1$ .
Rearrange: $6y + x - 31 = 0$ , option B.

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