Question

find parametrization for the curve of the lower half of the parabola x-1=y^2

Answer 1

is it x=t, and y=sqrt(t-1)

Answer 2

please explain

Answer 3

You can't find "the" parameterization of that curve, since there are many possible parameterizations. One possibility is to let y=t. Since x-1=y², we have x=y²+1, so x=t²+1 since y=t.

We've written x and y in terms of t, so now we just need to worry about the restrictions on t. Since the parabola x-1=y² opens horizontally and the coefficient of y is 0, we know that the axis of symmetry is the x-axis (if this is confusing, you could just graph the parabola instead). Thus, the lower half of the parabola (which is what we want to parameterize) is the part of the graph where y<0, i.e., where t<0.

So, one parameterization is x = t² + 1, y = t, t < 0.

Hope that helps! :)

Answer 4

thank you so much!