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#.
