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Question

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

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Solution

Given equation of curve is
y=x3+2x+6 .....(1)
Differentiating w.r.t x, we get
dydx=3x2+2

Let P(x1,y1) be any point on the curve.
Slope of the tangent to the given curve at P(x1,y1) is
(dydx)(x1,y1)=3x21+2

Slope of the normal to the given curve at point P(x1,y1) is
1Slope of the tangent at the point(x1,y1)
=13x2+2

The equation of the given line is x+14y+4=0 which can be written as
y=114x414 (which is of the form y=mx+c)
So, Slope of this line = 114
Since, the normal is parallel to this line.
So, slope of normal = slope of the given line.
13x21+2=114
3x21+2=14
3x21=12
x21=4
x1=±2
Since, P(x1,y1) lies on the curve
y1=x31+2x1+6 (by (1))
When x1=2,
y1=8+4+6=18

When x1=2,
y1=84+6=6
Therefore, there are two normals to the given curve with slope 114 and passing through the points (2,18) and (2,6).
Thus, the equation of the normal through (2,18) is given by,
y18=114(x2)
14y252=x+2
x+14y254=0
And, the equation of the normal through (2,6) is given by,
y(6)=114[x(2)]
y+6=114(x+2)
14y+84=x2
x+14y+86=0
Hence, the equations of the normals to the given curve (which are parallel to the given line) arex+14y254=0 and x+14y+86=0.

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