How to find a line normal to the curve $ xy = 1 $ ?

Hint: In the given question, we are required to describe the method of finding the equation of normal to the curve given to us as $ xy = 1 $ . To solve this type of question, let us suppose a general two dimensional line and equate its slope with the slope of the normal of the given curve which can be easily found out by differentiating the equation of the curve given to us. Then, we find the relation between coefficients of x and y of the supposed line.

Complete step by step solution:
So, let us suppose a two dimensional line in slope and intercept form $ y = mx + c $ .
So, the slope of line $ = m $
Now for the curve given to us $ xy = 1 $ ,
Differentiating both sides of the equation of the curve with respect to x,
We get,
 \[x\dfrac{{dy}}{{dx}} + y = 0\]
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{y}{x}\]
We can also find the $ \dfrac{{dy}}{{dx}} $ in only x using $ y = \dfrac{1}{x} $ .
So, we get, \[\dfrac{{dy}}{{dx}} = - \dfrac{1}{{{x^2}}}\]
So, this is the slope of tangent of the given curve $ xy = 1 $ .
As we know that the product of slope of tangent line and slope of normal line is always equal to $ - 1 $ . Then, the slope of normal at any point $ \left( {{x_1},{y_1}} \right) $ is \[{x_1}^2\] .
Now equating the slope of normal lines,
So, $ m = {x_1}^2 $ .
So, we get the equation of the line as $ y = {\left( {{x_1}} \right)^2}x + c $ .
Now, we can find the value of c by substituting the point $ \left( {{x_1},{y_1}} \right) $ that lies on the normal as well as the curve.
So, the correct answer is “ $ y = {\left( {{x_1}} \right)^2}x + c $ ”.

Note: \[\left( {\dfrac{{dy}}{{dx}}} \right)\] of any curve always represent slope of tangent line of the curve while \[ - \left( {\dfrac{{dx}}{{dy}}} \right)\] always represent the slope of normal line of the curve. This is because the product of the slopes of the tangent line and the normal line is always \[ - 1\] .