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Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 3 Sub Question 6 Maths Textbook Solution.

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ANSWER: Equation of tangent, 10 x-y-8=0

                 Equation  of normal, x+10 y-223=0


 Differentiating the given curve and find the slope.


y=x^{2}+4 x+1 \text { at } x=3


\frac{d y}{d x}=2 x+4

When x=3, y=9+12+1=22

So, \left(x_{1}, y_{1}\right)=(3,22)

Slope of tangent ,

m=\left(\frac{d y}{d x}\right)_{(x=3)}=10

Equation of tangent is,

\begin{aligned} &y-y_{1}=m\left(x-x_{1}\right) \\ &\Rightarrow y-22=10(x-3) \\ &\Rightarrow y-22=10 x-30 \quad \therefore 10 x-y-8=0 \end{aligned}

Equation of Normal  is,

\begin{aligned} &\Rightarrow y-y_{1}=-\frac{1}{m}\left(x-x_{1}\right) \\ &\Rightarrow y-22=-\frac{1}{10}(x-3) \\ &\Rightarrow 10 y-220=-x+3 \\ &\therefore x+10 y-223=0 \end{aligned}


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