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

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Answer 1 · 2016-08-02T15:54:17

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

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

For, $f(x)=-e^x, f'(x)=-e^x rArr f'(-2)=-e^-2=-1/e^2$ .

$:.$ the slope of tgt. to $C$ at $x=-2$ is $-1/e^2$ , &, since, normal is

$bot$ to tgt., the slope normal at $x=-2$ will be $-1/(-1/e^2)=e^2$

Also, $x=-2 rArr f(-2)=-e^-2=-1/e^2$ , so, the pt. of contact is $(-2,-1/e^2)$

Altogether, the normal passes thro. pt. $(-2,-1/e^2)$ and has slope

$=e^2$ . This gives us the eqn. of normal as $: y+1/e^2=e^2(x+2)$ ,

or, $ye^2+1=xe^4+2e^4, i.e., xe^4-ye^2+2e^4-1=0$ .

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