Answer

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**Hint:**

In this question, we’ll differentiate the curve to get the slope of the tangent then we’ll use $m \times n= -1$ where, $m$ and $n$ are the slope of tangent and normal respectively. A point on the given curve will satisfy both the equations. we’ll solve them to get the answer.

**Complete step by step solution:**

Let the point where normal is drawn be $\left( {{x}_{1}},\ {{y}_{1}} \right)$.

Give equation of curve at $xy=1$

So $xy=1$

Now differentiate with respect to x

$\dfrac{d\left( xy \right)}{dx}=\dfrac{d\left( 1 \right)}{dx}$

$\Rightarrow x\ \centerdot \ \dfrac{dy}{dx}+y=0\ \left[ \because \dfrac{d\left( \text{constant} \right)}{dx}=0 \right]$ $\left[ {by\;using\;product\;rule,\dfrac{d}{{dx}}\left( {xy} \right) = x\dfrac{d}{{dx}}\left( y \right) + y\dfrac{d}{{dx}}\left( x \right)} \right]$

$x\ \centerdot \ \dfrac{dy}{dx}=-y$

Therefore $\dfrac{dy}{dx}=\dfrac{-y}{x}$

Slope of tangent at \[P\left( {{x}_{1}},\ {{y}_{1}} \right)= \left( \dfrac{dy}{dx} \right)_{\left({x_1},{y_1}\right)}\]

Now $xy=1$

$y=\dfrac{1}{x}$$\Rightarrow y={{x}^{-1}}$

Differentiating with respect to x

$\dfrac{dy}{dx}=\dfrac{-1}{{{x}^{2}}}$ $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$

Slope of tangent at \[P\left( {{x}_{1}},{{y}_{1}} \right)=\left( \dfrac{dy}{dx} \right)_{\left({x_1},{y_1}\right)}=\dfrac{-1}{{{x}_{1}}^{2}}\]

\[\Rightarrow \dfrac{-1}{{{x}_{1}}^{2}}\times \text{Slope of normal=}\ -\text{1}\]

Therefore Slope of normal $={{x}^{2}}$ --(1)

Given equation of normal at $\left( {{x}_{1}},\ {{y}_{1}} \right)$is $ax+by+c=0$

Slope of normal is \[\dfrac{-a}{b}\] ---(2)

Now we can compare both the questions as they both are slopes of normal.

So from (1) and (2)

${{x}^{2}}=\dfrac{-a}{b}$

Since ${{x}^{2}}>0$

$

\Rightarrow \dfrac{-a}{b}>0 \\

\Rightarrow \dfrac{a}{b}<0 $

i.e. $\dfrac{a}{b}$ should be negative.

Therefore a <0, b>0 or a>0, b<0

**So option (B) and (C) are correct.**

**Note:**

There are two general forms of a line.

(1) $ax+by+c$

Slope of tangent$=\dfrac{-b}{a}$

Slope of normal$=\dfrac{-a}{b}$

(2) $y=mx+c$

Slope of tangent$=m$

Slope of normal$=\dfrac{-1}{m}$

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