Q. 214.0( 26 Votes )

Answer :

It is given that the equation of the normal to the curve y = x^{3} + 2x + 6

Then, the slope of the tangent to the given curve at any point (x, y) is given by:

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

=Mediumy =

⇒ Slope of the given line =

We know that 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.

⇒

⇒ 3x2 + 2 = 14

⇒ 3x^{2} = 12

⇒ x^{2} = 4

⇒

So, when x = 2 then y = 18

and x -2 then y = -6

Hence, there are two normals to the given curve with the slope and passing through the points (2, 18) and (-2,-6).

Then, the equation of the normal through (2, 18) is:

y -18 =

⇒ 14y -252 = -x +2

⇒ x +14y-254 = 0

And, the equation of the normal through (-2, -6) is:

y –(-6) =

⇒ y + 6 =

⇒ 14y +84 = -x -2

⇒ x +14y+86 = 0

Therefore, the equation of the normals to the curve y = x^{3} + 2x + 6 which are parallel to the line x + 14y + 4 = 0 are x +14y-254 = 0 and x +14y+86 = 0.

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