# the points of intersection of the curves whose parametric equations are x=t^2+1 , y=2t and x=2s , y=2/s is given by ?

And y^2 = 4 t^2

So we have 4 x = y^2 + 4 --(1)

By the second set, xy = 4--(2)

Now (1) * y gives

4 x y = y^3 + 4 y

Plug from (2)

16 = y^3 + 4 y

Or y^3 + 4 y - 16 = 0

==> y = 2

So x = 2

**(2,2)**is the point of intersection