Find the equation of the normals to the curve y = x3 + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
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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m×n = -1