Question

Find the point on the curve $y = {x^2} - 4x + 3$ , the normal at which is parallel to the y-axis. Also, find the equation of the normal.

Answer 1

Hint:
Firstly, find the slope of tangent to the curve $y = {x^2} - 4x + 3$ by differentiating the equation of the curve with respect to x. Then the slope of normal is the negative reciprocal of the slope of the tangent.
Thus, find the slope of normal to the curve.
Now, the point where the normal to the curve is parallel to the y-axis can be given by comparing the slopes of the normal and the slope of the y-axis.
Thus, we get the point.
Also, the equation of normal here can be given by \[x = a\] . So, find the equation of the normal to the curve $y = {x^2} - 4x + 3$.

Complete step by step solution:
It is given that a normal to the curve $y = {x^2} - 4x + 3$ is parallel to the y-axis.
To find the slope of normal, first we have to find the slope of the tangent to the curve $y = {x^2} - 4x + 3$ .
Now, we will differentiate the equation $y = {x^2} - 4x + 3$ with respect to x to get the slope of the tangent.
 $\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^2} - 4x + 3} \right)$
           $
= \dfrac{d}{{dx}}\left( {{x^2}} \right) - 4\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( 3 \right) \\
= 2x - 4 \\
 $
Thus, we get the slope of the tangent as $\dfrac{{dy}}{{dx}} = 2x - 4$ .
Now, slope of normal at that point is given by $m = - \dfrac{1}{{\dfrac{{dy}}{{dx}}}} = - \dfrac{1}{{2x - 4}}$ .
As it is given that, the normal at a point is parallel to the y-axis.
So, the slope of normal is the same as the slope of the y-axis. $\Rightarrow m = \dfrac{1}{0}$
 $
  \Rightarrow \dfrac{{ - 1}}{{2x - 4}} = \dfrac{1}{0} \\
  \Rightarrow - 1\left( 0 \right) = 2x - 4 \\
  \Rightarrow 2x - 4 = 0 \\
  \Rightarrow 2x = 4 \\
  \Rightarrow x = \dfrac{4}{2} \\
  \Rightarrow x = 2 \\
 $
Thus, \[x = 2\] . Also, to find the y co-ordinate of the point, substitute x = 2 in the equation of the curve $y = {x^2} - 4x + 3$ .
 $\Rightarrow y = {\left( 2 \right)^2} - 4\left( 2 \right) + 3$
$ = 4 - 8 + 3$
$=-1$
Thus, we get the point $\left( {2, - 1} \right)$ at which the normal to the curve $y = {x^2} - 4x + 3$ is parallel to the y-axis.
Now, the equation of a line parallel to the y-axis is given by \[x = a\] .
So, the equation of normal to the curve $y = {x^2} - 4x + 3$ which is parallel to the y-axis is \[x = 2\] .

Note:
If any line with slope m is parallel to the y-axis, then the slope of the line is the same as the slope of the y-axis.
 $\Rightarrow m = \dfrac{1}{0}$ , where m is the slope of the line and slope of y-axis is not defined i.e. $\dfrac{1}{0}$.
Also, the equation of the line parallel to the y-axis is given by $x = \pm a$ , where a is the distance of the line on either side of the y-axis.