You mean minus the reciprocal of the gradientOriginally posted by Teoh
Just differentiate the equation of the normal hyperbola to find gradient of tangent
Reciprocate that, and use two point formula with cartesian or paramteric co-ordinates to give you equation.
I can neither confirm, nor deny that comment...:mad1:Originally posted by Heinz
You mean minus the reciprocal of the gradient
Erm well you've done something wrong because thats wrong... try doing it again. You should end up with an x1^2/a^2 and a y1^2/b^2, which when you subtract the second from the first = 1Originally posted by edd91
parametric form
I know the process to get to it but I just cant??I get to this:
-xy1/a^2 + yx1/b^2= -yx/b^2 - xy/a^2
then dont know how to get to my target
english would be niceOriginally posted by Affinity
you can acquiesce