On a point #(ct,c/t)# on hyperbola #xy=c^2#, a normal is drawn which intersects the hyperbola at #(ct',c/(t'))#. Prove that #t^3t'=-1#?

Question

9077 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-28T07:28:57

Please see below.

The equation of hyperbola is #xy=c^2# and point #(ct,c/t)# lies on it.

Let us find the equation of the normal.

Equation of hyprbola can be written as #y=c^2/x# and therefore slope of tangent is given by first derivative i.e. #(dy)/(dx)=-c^2/x^2#

hence slope of normal is given by #x^2/c^2# and at #(ct,c/t)# is #t^2# and its equation is

#y=t^2(x-ct)+c/t#

or #xt^3-yt-ct^4+c=0#

As this passes through #(ct',c/(t'))#

#ct't^3-c/(t')t-ct^4+c=0#

or #ct^3(t'-t)+c/(t')(t'-t)=0#

as #t!=t'#, #t-t'!=0# and dividing by it we get

#ct^3=-c/(t')#

or #t^3t'=-1#