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

1 Answer
Aug 2, 2016

xe^4-ye^2+2e^4-1=0xe4ye2+2e41=0.

Explanation:

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.