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Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 10 Maths Textbook Solution.

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ANSWER: Equation of normal: y-18=\left(-\frac{1}{14}\right)(x-2) \text { or } y+6=\left(-\frac{1}{14}\right)(x-2)


 Differentiate  the given curve, to get slope of tangent


y=x^{3}+2x+6 Which is parallel to x+14y+4=0


Upon differentiation

\frac{dy}{dx}=3x^{2}+2\Rightarrow m(tangent)=3x^{2}+2

The normal is perpendicular to tangent, therefore , m_{1},m_{2}=-1

m(normal) at -\frac{1}{3x^{2}+2}

The equation of Normal  is given by  ,y-y_{1}=m\left ( tangent \right )\left ( x-x_{1} \right )

On comparing the slope of normal with the given equation


-\frac{1}{14}=-\frac{1}{\left ( 3x^{2}+2 \right )}\Rightarrow x=2\; or\; -2

 Thus ,the corresponding value of y is 18 or -6

Therefore ,the equation of normal are

y-18=\left(-\frac{1}{14}\right)(x-2) \text { or } y+6=\left(-\frac{1}{14}\right)(x+2)


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