Question

The equation of normal to the curve $y = {\log _e}^x$ at the point $P(1,0)$ is
$A)2x + y = 2 \\\ B)x - 2y = 1 \\\ C)x - y = 1 \\\ D)x + y = 1 \\\$

Answer 1

Hint: Here to find the equation of normal to the curve given let’s find out the slope of normal (given) and then use the slope point form of the equation of line.

Complete step-by-step answer:
Here we have to find the equation of normal to given curve
We know that the normal and the tangent of a curve at a point on the curve are perpendicular
So,
Slope of normal = $\dfrac{{ - 1}}{{{\text{Slope of tangent}}}}$
Here to get the slope of the tangent lets differentiate the given curve with respective x.
On differentiating with respective x we get
$y'(x) = \dfrac{1}{x}\left[ {\because \dfrac{d}{{dx}}({{\log }_e}x) = \dfrac{1}{x}} \right]$
Now, substituting the point given $P(1,0)$ in the above equation we get,
$y'(1) = \dfrac{1}{1} = 1$.
Slope of normal = $\dfrac{{ - 1}}{{{\text{Slope of tangent}}}}$=$\dfrac{{ - 1}}{{y'(1)}} = \dfrac{{ - 1}}{1} = - 1$
Therefore, the slope of normal to the curve $y = {\log _e}^x$ at point $P(1,0)$ is $- 1$
And we also know that the normal passes through the point $P$
Formula for Equation of line =$y - {y_1} = m(x - {x_1})$
Then equation of normal that passes through point $P(1,0)$ with slope (m) = $- 1$
So to get the equation of normal let us substitute the value in the equation line formula. Where we get it as
$\Rightarrow y - 0 = - 1(x - 1) \\\ \Rightarrow x + y = 1 \\\$
Therefore equation of normal is x+y=1
Option D is the correct answer.

Note: We should note the point that the normal of the given curve passes through the given point P and also we have to know the relation between normal and tangent to solve.