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

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

Answers (1)

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

 

 

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