Find coordinates of a point for a derivative of a parametric curve

Question

Find the coordinates of the points at which the given parametric curve has a) a horizontal tangent and b) a vertical tangent.

The parametric curve is $$\mathscr{C}=\begin{cases}x=t^2+1\\y=2t-4\end{cases}$$

I took immediately the derivative with respect to t to get, $x=2t$ and $y=2$ to get the point of which the parametric curve has a horizontal tangent. So that would be $(2t,2)$. The slope is then $$y'/x'=2/2t=1/t$$. So for the slope to be zero (horizontal tangent), we would solve $$0=1/t$$ and get P=(0,1) where the tangent is horizontal.

But for the vertical tangent, I am not sure.

Any hints?

Thanks

Answer 1

Vertical tangents occur where $x’=0$, so at $t=0$.

Horizontal tangents occur where $y’=0$. Of course, this never happens, so there are no horizontal tangents.

All of this is to be expected, because, if you eliminate $t$ you find that $x$ is a quadratic function of $y$, and the curve is a parabola that opens to the right.