# What is the equation of the line that is normal to f(x)=-e^x  at  x=-2 ?

Aug 2, 2016

$x {e}^{4} - y {e}^{2} + 2 {e}^{4} - 1 = 0$.

#### Explanation:

Let us recall that, $f ' \left(x\right)$ is the slope of tgt. to the curve $C : y = f \left(x\right)$ at any pt. $\left(x , y\right)$.

For, $f \left(x\right) = - {e}^{x} , f ' \left(x\right) = - {e}^{x} \Rightarrow f ' \left(- 2\right) = - {e}^{-} 2 = - \frac{1}{e} ^ 2$.

$\therefore$ the slope of tgt. to $C$ at $x = - 2$ is $- \frac{1}{e} ^ 2$, &, since, normal is

$\bot$ to tgt., the slope normal at $x = - 2$ will be $- \frac{1}{- \frac{1}{e} ^ 2} = {e}^{2}$

Also, $x = - 2 \Rightarrow f \left(- 2\right) = - {e}^{-} 2 = - \frac{1}{e} ^ 2$, so, the pt. of contact is $\left(- 2 , - \frac{1}{e} ^ 2\right)$

Altogether, the normal passes thro. pt. $\left(- 2 , - \frac{1}{e} ^ 2\right)$ and has slope

$= {e}^{2}$. This gives us the eqn. of normal as $: y + \frac{1}{e} ^ 2 = {e}^{2} \left(x + 2\right)$,

or, $y {e}^{2} + 1 = x {e}^{4} + 2 {e}^{4} , i . e . , x {e}^{4} - y {e}^{2} + 2 {e}^{4} - 1 = 0$.