#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 11 Maths Textbook Solution.

ANSWER: Equation of tangent: $9x-y-3=0\; or\; 9x-y+13=0$

HINTS:

Slope of tangent = slope of perpendicular line

GIVEN:

$y=4x^{3}-3x+5$ Which is perpendicular to $9y+x+3=0$

SOLUTION:

Let $\left ( x_{1},y_{1} \right )$ be a point on the curve which are used to find the tangents

Slope of the given line = $-\frac{1}{9}$

Since tangent is perpendicular to the given line ,

Slope of the tangent= $\frac{-1}{\left ( -\frac{1}{9} \right )}=9$

Let $\left ( x_{1},y_{1} \right )$ be a point where the tangent is drawn to this curve

Since, the point lies on the curve

Hence,

$y_{1}=4x^{3}_{1}-3x_{1}+5$

Slope of the tangent =slope of the perpendicular line

\begin{aligned} &\Rightarrow 12 x_{1}^{2}-3=9 \quad \Rightarrow 12 x_{1}^{2}=12\\ &\Rightarrow x_{1}^{2}=1 \quad \Rightarrow x_{1}=\pm 1\\ &\text { Case- } 1: x_{1}=1\\ &y_{1}=4 x_{1}^{3}-3 x_{1}+5=4-3+5=6\\ &\therefore\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(1,6)\\ &\\&\text { Case- } 2: x_{1}=-1\\ &y_{1}=4 x_{1}^{3}-3 x_{1}+5=-4+3+5=4\\ &\therefore\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(-1,4) \end{aligned}

$\\\text{Thus, the equation of tangent is}\\ (y-6)=9(x-1) \ \ \ \Rightarrow 9 x-y-3=0 \\ (y-4) =9(x+1) \ \ \ \Rightarrow 9 x-y+13=0$