Find equations of normal and tangent lines at given points.
$x=2t-t^2$
$y=3t-t^3$
at $t=1$ point.
$\frac{dx}{dt}=2-2t$
$\frac{dy}{dt}=3-3t^2$
$\frac{dy}{dx}=\frac{3-3t^2}{2-2t}$
from here I don't know what to do slope at point $t=1$ becomes $0/0$.
Answer is $3x-y-1=0$ $x+3y-7=0$
Note that, if $t\ne1$, then$$\frac{y'(t)}{x'(t)}=\frac{3-3t^2}{2-2t}=\frac{3+3t}2.$$So, $\lim_{t\to1}\frac{y'(t)}{x'(t)}=3$, and therefore the tangent line when $t=1$ is the line with slope $3$ passing through $(1,2)$, which is indeed the line $3x−y−1=0$; see the graph in picture below: