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Question: If the line \(ax+by+c=0\) is normal to the curve \[xy\ \text{=}\ \text{1}\], then this question has ...

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

Explanation

Solution

In this question, we’ll differentiate the curve to get the slope of the tangent then we’ll use m×n=−1m \times n= -1 where, mm and nn 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 (x1, y1)\left( {{x}_{1}},\ {{y}_{1}} \right).
Give equation of curve at xy=1xy=1
So xy=1xy=1
Now differentiate with respect to x
d(xy)dx=d(1)dx\dfrac{d\left( xy \right)}{dx}=\dfrac{d\left( 1 \right)}{dx}
⇒x ⋅ dydx+y=0 [∵d(constant)dx=0]\Rightarrow x\ \centerdot \ \dfrac{dy}{dx}+y=0\ \left[ \because \dfrac{d\left( \text{constant} \right)}{dx}=0 \right] [by  using  product  rule,ddx(xy)=xddx(y)+yddx(x)]\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 ⋅ dydx=−yx\ \centerdot \ \dfrac{dy}{dx}=-y
Therefore dydx=−yx\dfrac{dy}{dx}=\dfrac{-y}{x}
Slope of tangent at P(x1, y1)=(dydx)(x1,y1)P\left( {{x}_{1}},\ {{y}_{1}} \right)= \left( \dfrac{dy}{dx} \right)_{\left({x_1},{y_1}\right)}
Now xy=1xy=1
y=1xy=\dfrac{1}{x} ⇒y=x−1\Rightarrow y={{x}^{-1}}
Differentiating with respect to x
dydx=−1x2\dfrac{dy}{dx}=\dfrac{-1}{{{x}^{2}}} d(xn)dx=nxn−1\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}
Slope of tangent at P(x1,y1)=(dydx)(x1,y1)=−1x12P\left( {{x}_{1}},{{y}_{1}} \right)=\left( \dfrac{dy}{dx} \right)_{\left({x_1},{y_1}\right)}=\dfrac{-1}{{{x}_{1}}^{2}}
⇒−1x12×Slope of normal= −1\Rightarrow \dfrac{-1}{{{x}_{1}}^{2}}\times \text{Slope of normal=}\ -\text{1}
Therefore Slope of normal =x2={{x}^{2}} --(1)
Given equation of normal at (x1, y1)\left( {{x}_{1}},\ {{y}_{1}} \right)is ax+by+c=0ax+by+c=0
Slope of normal is −ab\dfrac{-a}{b} ---(2)
Now we can compare both the questions as they both are slopes of normal.
So from (1) and (2)
x2=−ab{{x}^{2}}=\dfrac{-a}{b}
Since x2>0{{x}^{2}}>0
⇒−ab>0 ⇒ab<0 \Rightarrow \dfrac{-a}{b}>0 \\\ \Rightarrow \dfrac{a}{b}<0

i.e. ab\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+cax+by+c
Slope of tangent=−ba=\dfrac{-b}{a}
Slope of normal=−ab=\dfrac{-a}{b}
(2) y=mx+cy=mx+c
Slope of tangent=m=m
Slope of normal=−1m=\dfrac{-1}{m}