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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the equation of the normals to the curve \(y = x^3 + 2x + 6\) which are parallel to the line \(x + 14y + 4 = 0.\)

$\begin{array}{1 1} (A)\;x+14y-254=0;x+14y+86=0 \\ (B)\;x-14y-254=0;x+14y+86=0 \\(C)\;x+14y-254=0;x-14y+86=0 \\ (D)\;x-14y-254=0;x-14y-86=0 \end{array} $

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Toolbox:
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
Equation of the curve is $y=x^3+2x+6$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=$$3x^2+2$
Equation of the given line is $x+14y+4=0$
Slope of the given line is $-\bigg(\large\frac{coeff\;of\;x}{coeff of y}\bigg)$
$\Rightarrow \large\frac{-1}{14}$
Since the normal is parallel to the given line hence their slopes are equal.
(i.e)$\large\frac{-1}{14}=\frac{-1}{3x^2}$
Step 2:
$3x^2+2=14$
$3x^2=12$
$x^2=4$
$x=\pm 2$
Step 3:
When $x=2$,
$y=2^3+2(2)+6$
$y=16$
When $x=-2$
$y=(-2)^3+2(-2)+6$
$\;\;\;=-8-4+6$
$\;\;\;=-6$
Hence the points are $(2,16)$ and $(-2,-6)$
Step 4:
Hence the equations of the normals are $(y-16)=\large\frac{-1}{14}$$(x-2)$
$\Rightarrow x+14y-254=0$
$(y+4)=\large\frac{-1}{14}$$(x+2)$
$\Rightarrow x+14y+86=0$
answered Jul 11, 2013 by sreemathi.v
 

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