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

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

Hint:

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

Given:

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.

Solution:

Given,

\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  $\left ( -\infty,0 \right )$