# Prove that the curve $\alpha(t)$ is tangent to the $x$ axis.

I have the curve $$\alpha(t):(-1,\infty) \rightarrow R^2$$ given by $$\alpha(t)= ((\frac{nat}{1+t^3}), (\frac{nat^2}{1+t^3}))$$ with $$n$$ a natural an $$a$$ a constant both of them fixed. I need to prove that this curve is tangent to the $$x$$ axis in $$t=0$$.

What I did:

It's evident that the only point where the curve touch the $$x$$ axis is in t=0, so $$\alpha'(t) = (\frac{na(1+t^3 - nt^2(nat)}{(1+t^3)^2},\frac{2nat(1+t^3)-nt^2(nat^2)}{(1+t^3)^2})$$ and $$\alpha'(0) = (na,0)$$, so the tangent of the curve is parallel to the $$x$$ axis and then the curve is tangent to the $$x$$ axis in $$t=0$$.

It's this all right?

• Yes, that's right – Sebastian Schulz Mar 2 at 19:09