1nishant can u explain me how did u get the slope as t^2
general eqn of normal to rectangular hyperbola xy=c2
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3 Answers
62we got that the slope of the normal at ct,c/t is
t2
so eqtn will be
(y-c/t)=t2(x-ct)
62slope of tangent is dy/dx
xy=c2
y+x dy/dx=0
dy/dx = -y/x
substitute y=c/t ... x=ct
you will get slope of tangent =-1/t2
slope of normal is -1/m where m is the slope of tangent... hence t2
sorry for the late late late reply :) ;)