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  • Need solution for RD Sharma Maths Class 12 Chapter 15 Tangents and Normals Excercise Fill in the blanks Question 17

Need solution for RD Sharma Maths Class 12 Chapter 15 Tangents and Normals Excercise Fill in the blanks Question 17

Answers (1)

Answer:

Equation of normal,  y=3x-5

Hint:

Use equation of normal,

                \left(y-y_{1}\right)=-\frac{d x}{d y}\left(x-x_{1}\right)

Given:

Given curve,

                x^{2}+y^{2}-2 x+4 y-5=0

To find:

We have to find the slope of normal to the given curve at point  \left ( 2,1 \right )

Solution:

Given equation is

                x^{2}+y^{2}-2 x+4 y-5=0

On differentiating both side with respect to x , we get

\Rightarrow \quad 2 x+2 y \frac{d y}{d x}-2+4 \frac{d y}{d x}=0                                                                                         \left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]

\Rightarrow \quad(y+2) \frac{d y}{d x}=1-x     [Taking common 2  from each term]

\begin{aligned} &\Rightarrow \quad \frac{d y}{d x}=\frac{1-x}{y+2} \\\\ &\Rightarrow \quad\left(\frac{d y}{d x}\right)_{(2,1)}=\frac{1-2}{1+2}=\frac{-1}{3} \\ \end{aligned}

\Rightarrow \quad-\left(\frac{d y}{d x}\right)_{(2,1)}=\frac{1}{3}

\Rightarrow \quad-\left(\frac{d x}{d y}\right)_{(2,1)}=3

Now, equation of normal is

              \left(y-y_{1}\right)=-\frac{d x}{d y}\left(x-x_{1}\right)

Substituting these value, we get

           i.e.   y_{1}=1, x_{1}=2, \frac{-d x}{d y}=3

             \begin{aligned} &(y-1)=3(x-2) \\ \Rightarrow & \quad y-1=3 x-6 \\ \Rightarrow & \quad y=3 x-5 \end{aligned}

Hence, y=3 x-5 is required equation of normal.

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