Answer
Point of horizontal tangency: (4,0)
Work Step by Step
The point of horizontal tangency will occur when $dy/dx = 0$, therefore when $dy$ is zero, in this case:
$y = t^2$
$dy = 2tdt$
dt can't be 0, so: $2t = 0$: $t = 0$
Substitute the values at t = 0:
$x = 4 - t = 4 - 0 = 4$
$y = t^2 = (0)^2 = 0$
Therefore, the point of horizontal tangency is (4,0)
The point of vertical tangency will occur when $dx$ is equal to 0:
$x = 4-t$
$dx = -1dt$
-1dt can't be equal to 0, so there is no vertical tangency point.
