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2075 Points
one year ago
let the 2 fns be y= f(x) and y= g(x). now consider a point with abscissa x. then eqn of tangent for f and g would be:
Y – f(x)= f’(x)(X – x)
and Y – g(x)= g’(x)(X – x). here X and Y represent the axes.
now, since they intersect at y axis, X=0 so that
Y= f(x) – xf’(x)= g(x) – xg’(x).....(1)
similarly the eqn of normals to f and g would be:
Y – f(x)= (–1/f’(x))(X – x)
and Y – g(x)= (–1/g’(x))(X – x). these intesect on x axis so that Y=0
so X= x + f(x)*f’(x)= x + g(x)*g’(x)
or f(x)*f’(x)= g(x)*g’(x).....(2)
from (2), we can say
f(x)*df(x)/dx= g(x)*dg(x)/dx
integrating both sides,
∫f(x)df(x)= ∫g(x)dg(x)
f^2(x)/2= g^2(x)/2 + C
or f^2(x)= g^2(x) + K......(3) where K= 2C is a constant.
(1) can be written as
f(x) – g(x)= x(f’(x) – g’(x))
f(x) – g(x)= xd(f(x) – g(x))/dx
or dx/x= d(f(x) – g(x))/f(x) – g(x)
integrate
ln|x|= ln|f(x) – g(x)| + C
raise to power of e
x= A(f(x) – g(x)), ….(4) A= ±e^C being a constant.
(3) and (4) can be solved for f and g by noting that (3) can be factorised as (f – g)(f+g)= K and then putting f – g= x/A from (4) and finally adding these eqns.
we shall obtain
f(x)= (KA/x + x/A)/2 and g(x)= (KA/x – x/A)/2
if we let 2A= B (another constant) and K= 4M, then
f(x)= MB/x + x/B and g(x)= MB/x – x/B here M and B are constants and obviously B cant be zero.
kindly approve :)