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

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

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