A curve C is defined by the parametric equations: #x=t^2# and #y = t^3-3t#, how do you show that C has two tangents at the point (3,0) and find their equations?

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Answer 1 · 2016-12-19T17:19:52

See below

#dy/dx = (dy/(dt))/(dx/(dt)) = (3t^2-3)/(2t)#

#x=3# at #t = +-sqrt3# so there are two slopes for tangent lines

#m_1 = (3(sqrt3)^2-3)/(2sqrt3) = 6/(2sqrt3) = sqrt3#

and

#m_2 = (3(-sqrt3)^2-3)/(2(-sqrt3)) = 6/(-2sqrt3) = -sqrt3#.

The equations are found by finding the equations of the lines through #(3,0)# with slopes #sqrt3# and #-sqrt3#.

#y=sqrt3(x-3)# and #y=-sqrt3(x-3)#