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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.

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

Answers (1)

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

HINTS:

 Differentiate  the given curve, to get slope of tangent

GIVEN:

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

SOLUTION:

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

m(normal)=-\frac{1}{14}

-\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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