Question

The equation of normal to the curve y = 3x² − x + 1 at (1, 3) is ______.

Options

• x − 5y − 16 = 0

• x + 5y − 16 = 0

• x − 5y + 16 = 0

• −5y − x − 16 = 0

Answer 1

The equation of normal to the curve y = 3x² − x + 1 at (1, 3) is x + 5y − 16 = 0.

Explanation:

Differentiating w.r.t. x, we get

`dy/dx = d/dx (3x^2 − x + 1)`

`dy/dx = 6x - 1`

Slope of the tangent at (1, 3) is

`("dy"/"dx")_((1,3)` = 6(1) − 1 = 5

The normal line is perpendicular to the tangent line. The relationship between their slopes is `m_n = - 1/m_t` Therefore:

`m_n = -1/5`

∴ Equation of the normal at (1, 3) is

`y - 3 = -(1)/5(x - 1)`

`5(y -3) = -1(x - 1)`

∴ 5y − 15 = − x + 1

∴ x + 5y − 16 = 0