On the curve x=2t³-9t²+12t-1, y=t²+t+1, find the points
On the curve $x=2t^{3}-9t^{2}+12t-1,\ y=t^{2}+t+1$, find the points at which the tangent is parallel to the $Oy$ axis.
Explanation of Method
A vertical tangent is where dx/dt vanishes while dy/dt does not.
Step 1: Condition for a vertical tangent
The tangent is vertical exactly where $\dot x = 0$ and $\dot y\neq0$.
Step 2: Derivatives
\[ \dot x = 6t^{2}-18t+12 = 6(t-1)(t-2), \qquad \dot y = 2t+1. \tag{1} \]
Step 3: Solve $\dot x=0$
\[ t=1 \ \text{ or } \ t=2, \qquad \dot y(1)=3\neq0,\ \ \dot y(2)=5\neq0. \tag{2} \]
Both give genuine vertical tangents.
Step 4: The corresponding points
\[ t=1\colon\ x=2-9+12-1=4,\ y=3 \;\Longrightarrow\; P_{1}=(4,3). \]
\[ t=2\colon\ x=16-36+24-1=3,\ y=7 \;\Longrightarrow\; P_{2}=(3,7). \]
Step 5: Verification
\[\dot x(1)=6(0)(-1)=0\ \checkmark;\qquad \dot x(2)=6(1)(0)=0\ \checkmark\]
Result
Final answer:
\[\boxed{\,P_{1}=(4,3)\ \text{(at } t=1\text{)} \qquad \text{and} \qquad P_{2}=(3,7)\ \text{(at } t=2\text{)}\,}\]