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

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Equation of normal,  y=3x-5


Use equation of normal,

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


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 )


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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