im getting the answer f(x)=2e^-x/2....

solution..

slope of the tangent (m) = f '(p)

the eqn of tangent at [p,f(p)]=>  

y-f(p)=f '(p) (x-p).... it passes trough [(p+2),0]

so.. -f(p)=f '(p)(2)

as p is any point at the curve we can replace p by x and f(p) by f(x) that is.. y.

 

so we get:

-y=2dy/dx

-2(dy)/y=dx

integrating..

 

-2lny=x+c.. now this is the original curve which passes through (0,2) so, we get the equation by putting the value of c...

-2lny=x-2ln2