# How do you find the equation of the tangent and normal line to the curve y=lnx at x=e?

Sep 9, 2016

Tangent: $y = \frac{x}{e}$
Normal: $y = e \left(e - x\right) + 1$

#### Explanation:

$y = \ln x$
$\therefore y = 1$ at $x = e$

$y ' = \frac{1}{x} \to$ the slope of $f \left(x\right) = \frac{1}{x}$

The equation of a straight line with slope $\left(m\right)$ passing throught point $\left({x}_{1} , {y}_{1}\right)$ is;

$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$

Since $m$ at $x = e$ is $\frac{1}{e}$

The equation of tangent to $f \left(x\right)$ at $\left(e , 1\right)$ is
$\left(y - 1\right) = \frac{1}{e} \left(x - e\right)$

$y = \frac{x}{e} - 1 + 1$

$y = \frac{x}{e}$

And equation of normal to $f \left(x\right)$ at $\left(e , 1\right)$ is
$\left(y - 1\right) = - e \left(x - e\right)$ (Since slope of normal $= - \frac{1}{m}$)

$y = - e x + {e}^{2} + 1$

$y = e \left(e - x\right) + 1$