Question

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

Answer 1

The equation of the given curve is y = x3 + 2x + 6. The slope of the tangent to the given curve at any point (x, y) is given by,

$\frac{dy}{dx}$=3x2+2

∴ The slope of the normal to the given curve at any point (x, y)

=$\frac{-1}{slope\, of \,the\,tangent\,at\,the \,point\,(x,y)}$

=$\frac{1}{3x^2+2}$

The equation of the given line is x + 14y + 4 = 0.

x + 14y + 4 = 0

∴ y=$-\frac{1}{14}x-\frac{4}{14}$ (which is of the form y = mx + c)

∴The slope of the given line = $-\frac{1}{14}$

If the normal is parallel to the line, then we must have the slope of the normal being equal to the slope of the line.

$-\frac{1}{3x^2+2}$=$-\frac{1}{14}$

3x2+2=14

3x2=12

x2=4

x=±2

When x = 2, y = 8 + 4 + 6 = 18.

When x = −2, y = − 8 − 4 + 6 = −6.

Therefore, there are two normals to the given curve with slope $-\frac{1}{14}$ and passing through the points (2, 18) and (−2, −6).

Thus, the equation of the normal through (2, 18) is given b

y-18=-$\frac{1}{14}$(x-2)

14y-252=-x+2

x+14y-254=0

And, the equation of the normal through (−2, −6) is given by,

y-(-6)=-$\frac{1}{14}$ (x+2)

14y+84=-x-2

x+14y+86=0

Hence, the equations of the normals to the given curve (which are parallel to the given line) are x+14y-254=0 and x+14y+86=0.