Find the equations of tangent and normal to the curve y = 3x^{2} - 3x - 5 where the tangent is parallel to the line 3x − y + 1 = 0.

#### Solution

Equation of the curve is y = 3x^{2} - 3x - 5

Differentiating w.r.t. x, we get

`"dy"/"dx"` = 6x - 3

Slope of the tangent at P(x_{1}, y_{1}) is

`("dy"/"dx")_(("x"_1,"x"_2)` = 6x_{1} - 3

According to the given condition, the tangent is parallel to 3x – y + 1 = 0

Now, slope of the line 3x – y + 1 = 0 is 3.

∴ Slope of the tangent = `"dy"/"dx" = 3`

∴ 6x_{1} - 3 = 3

∴ x_{1} = 1

P(x_{1}, y_{1}) lies on the curve y = 3x^{2} - 3x - 5

∴ y_{1 }= 3(1)^{2} - 3(1) - 5

∴ y_{1 }= - 5

∴ The point on the curve is (1, -5).

∴ Equation of the tangent at (1, -5) is

∴ (y + 5) = 3(x – 1)

∴ y + 5 = 3x – 3

∴ 3x - y - 8 = 0

Slope of the normal at (1, -5) is `(-1)/("dy"/"dx")_((1,-5)` = `(- 1)/3`

∴ Equation of the normal of (1, -5) is

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

∴ 3y + 15 = - x + 1

∴ x + 3y + 14 = 0