For intuition: Imagine that the function #f(x,y) = e^x ln(y) - xy# describes the height of some terrain, where #x# and #y# are coordinates in the plane and #ln(y)# is assumed to be the natural logarithm. Then all #(x,y)# such that #f(x,y)=a# (the height) equals some constant #a# are called level curves . In our case the constant height #a# is zero, since #f(x,y)=0#.
You might be familiar with topographic maps, in which the closed lines indicate lines of equal height.
Now the gradient #grad f (x,y)= ( (partial f) /(partial x), (partial f) /(partial x) ) = (e^x ln(y) - y, e^x/y - x)# gives us the direction at a point #(x,y)# in which #f(x,y)# (the height) changes the fastest. This is either straight up or straight down the hill, as long as our terrain is smooth (differentiable), and we are not on a top, in a bottom or on a plateau (an extremum point). This is in fact the normal direction to a curve of constant height, such that at #(x,y)=(2,e^2)#:
#grad f (2,e^2) = (e^2 ln(e^2) - e^2, e^2/e^2 - 2)=(e^2,-1)#.
Therefore, the normal line in that direction going through #(2,e^2)# can be described as
#(x,y) = (2,e^2) + s(e^2,-1)#,
where #s in mathbbR# is a real parameter. You can eliminate #s# to express #y# as a function of #x# if you prefer, to find
#y=(x-2-e^4)/e^2#.
The directional derivative in the tangent direction must be #0# (meaning that height does not change), so a tangent vector #(u,v)# must satisfy
#grad f(2, e^2)cdot (u,v)=0#
#(e^2,-1) cdot (u,v) = 0#
#e^2u - v=0#
#v=e^2u#,
where #cdot# means the dot product. So #(u,v) = (1,e^2)# is one valid choice. Therefore, the tangent line going through #(2,e^2)# can be described as
#(x,y) = (2,e^2) + t(1,e^2)#, #t in mathbbR#.
Solving for #y# gives that
#y = e^2x -e^2#.
You should finally check that #(2,e^2)# lies on the curve #f(x,y)#, on the tangent line, and on the normal line.
