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  • Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 4 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 4 Maths Textbook Solution.

Answers (1)

Answer: Equation of tangent, x-y-3=0

           Equation of normal, x+y+1=0

HINTS:

 Differentiating the given equation with respect to x.

Given:

y=2 x^{2}-3 x-1 \text { at }(1,-2)

Solution:

\frac{d y}{d x}=4 x-3

Slope of tangent,

m=\left(\frac{d y}{d x}\right)_{|1,-2|}=4-3=1

Equation of tangent is,

\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &\Rightarrow y+2=1(x-1) \\ &\Rightarrow y+2=x-1 \quad \therefore x-y-3=0 \end{aligned}

Equation of Normal  is,

\begin{aligned} &\Rightarrow y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) \\ &\Rightarrow y+2=-1(x-1) \\ &\Rightarrow y+2=-x+1 \\ &\Rightarrow x+y+1=0 \end{aligned}

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