Find the equations of the normals to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

First, we need to find normals to the curve y = x³ + 2x + 6, and then find the equations of those normals.

Given curve, y = x³ + 2x + 6

Let (h, k) be the point on the curve.

We shall find the normal at this point on the curve.

So, (h, k) will satisfy equation of the given curve, such that

(x, y) ≡ (h, k)

This means,

k = h³ + 2h + 6 …(i)

We know that, slope of the tangent of any curve is given by .

Take the equation of the curve,

y = x³ + 2x + 6

Differentiate with respect to x,

Since, the tangent is to be taken from point (h, k).

Slope of tangent at (h, k) is,

⇒ Slope of tangent = 3h² + 2

We know the relationship between the slope of tangent and slope of normal, that is,

Slope of tangent × Slope of normal = -1

⇒ (3h² + 2) × Slope of normal = -1

Also, given that curve is parallel to the line x + 14y + 4 = 0.

This means, normal of the curve is also parallel to the line x + 14y + 4 = 0.

And, if two lines are parallel, then their slopes are equal.

That is, slope of normal = slope of the line (x + 14y + 4 = 0) …(ii)

Take line x + 14y + 4 = 0

⇒ 14y = -x – 4

Let us rearrange the above equation in the form y = mx + c, where m is the slope.

Here, .

From (ii),

slope of normal = slope of the line (x + 14y + 4 = 0)

By cross-multiplication, we get

⇒ 3h² + 2 = 14

⇒ 3h² = 14 – 2

⇒ 3h² = 12

⇒ h² = 4

⇒ h = ±√4

⇒ h = ±2

Put this value of h in equation (i).

When h = 2,

k = h³ + 2h + 6

⇒ k = (2)³ + 2(2) + 6

⇒ k = 8 + 4 + 6

⇒ k = 18

We get the point (h, k) ≡ (2, 18).

When h = -2,

k = h³ + 2h + 6

⇒ k = (-2)³ + 2(-2) + 6

⇒ k = -8 – 4 + 6

⇒ k = -12 + 6

⇒ k = -6

We get the another point (h, k) ≡ (-2, -6).

Now, as we have found out points on the curve, we shall proceed to find the equation of the normal.

Equation of line at point (x₁, y₁) and having slope m is

y – y₁ = m(x – x₁)

Let us find equation of the line at point (2, 18) having slope -1/14 is,

Solving it further,

⇒ 14(y – 18) = -(x – 2)

⇒ 14y – 252 = -x + 2

⇒ 14y + x = 252 + 2

⇒ 14y + x = 254

⇒ x + 14y – 254 = 0

And also, equation of the line at point (-2, -6) having slope -1/14 is,

Solving it further,

⇒ 14(y + 6) = -(x + 2)

⇒ 14y + 84 = -x – 2

⇒ x + 14y + 84 + 2 = 0

⇒ x + 14y + 86 = 0

Thus, the equations are x + 14y – 254 = 0 and x + 14y + 86 = 0.