Question:

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

Updated On: Sep 15, 2023
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Solution and Explanation

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.

 

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Concepts Used:

Tangents and Normals

  • A tangent at a degree on the curve could be a straight line that touches the curve at that time and whose slope is up to the derivative of the curve at that point. From the definition, you'll be able to deduce the way to realize the equation of the tangent to the curve at any point.
  • Given a function y = f(x), the equation of the tangent for this curve at x = x0 
  • Slope of tangent (at x=x0) m=dy/dx||x=x0
  • A normal at a degree on the curve is a line that intersects the curve at that time and is perpendicular to the tangent at that point. If its slope is given by n, and also the slope of the tangent at that point or the value of the derivative at that point is given by m. then we got 

m×n = -1

  • The normal to a given curve y = f(x) at a point x = x0
  • The slope ‘n’ of the normal: As the normal is perpendicular to the tangent, we have: n=-1/m

Diagram Explaining Tangents and Normal: