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

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                \frac{a}{b}   Should be negative. i.e.\left ( -\infty,0 \right )


First find the slope of normal to the given curve then compare with slope of  a x+b y+c


Given the equation of curve,

                xy= 1

To find:

We have to find the set of \frac{a}{b}   if the line  a x+b y+c=0  is normal to the given curve.



                \begin{aligned} & x y=1 \\ \Rightarrow \quad & y=\frac{1}{x} \end{aligned}

Differentiating with respect to x , we get

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

Thus slope of normal  =\frac{-1}{\frac{d y}{d x}}=x^{2}

Which is always positive and it is given  a x+b y+c=0  is normal

Slope  =\frac{-a}{b}

\Rightarrow \quad \frac{-a}{b}>0

\Rightarrow \frac{a}{b}<0

So a and b are of opposite sign.

Hence   a<0, b>0

=> \frac{a}{b} should be negative. i.e  \left ( -\infty,0 \right )

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