Question

The equation of normal to the curve $y=lo{g}_e^x$ 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

The correct option is D $x+y=1$

To find the equation for the normal to the curve $y=lo{g}_{e}x$ at the point $P(1,0)$, we find the slope of said normal and then use the slope-point form of the equation of a line.

We know that the normal and the tangent of a curve at a point on the curve are perpendicular. So,

$\text{Slope of the normal}=\frac{-1}{\text{Slope of the tangent}}=\frac{-1}{y^{\prime }\left(1\right)}$.

Now, $y^{\prime }\left(x\right)=\frac{1}{x}$. So, $y^{\prime }\left(1\right)=1$.

Therefore, the slope of the normal to the curve $y=lo{g}_{e}x$ at the point $P(1,0)$ is $-1$.

Noting that, the normal passes through the point $P$, we find its equation as:

$y-0=-1\times (x-1)$.

That is, $x+y=1$.