Find out the x-y coordinates of the points in which the following parametric equations will have horizontal or vertical tangents.

x = t³ - 3t       

y = 3t² - 9

Solution

We'll first require the derivatives of the parametric equations.

dx/dt = 3t² - 3 = 3 (t² -1)                                            dy/dt = 6t

Horizontal Tangents

We'll have horizontal tangents in which,

6t = 0   ⇒ t = 0

Now here, this is the value of t that gives the horizontal tangents and we were asked to find out the x-y coordinates of the point. To get these we just only need to plug t into the parametric equations. Hence, the just only horizontal tangent will take place at the point (0,-9).

Vertical Tangents

In this case we require to solve,

3(t² -1) = 0      ⇒      t = ≠1

The two vertical tangents will take place at the points (2,-6) and (-2,-6). On behalf of completeness and at least partial verification here is the drawing of the parametric curve.