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# If the line $ax+by+c=0$ is normal to the curve $xy\ \text{=}\ \text{1}$, then this question has multiple correct options.a) $a>0,\ b>0$ b) $a>0,\ b<0$ c) $a<0,\ b>0$ d) $a<0,\ b<0$

Last updated date: 02nd Aug 2024
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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$=\dfrac{-1}{m}$