Find the equations of the tangents to the curve x=t²

Find the equations of the tangents to the curve $x=t^{2},\ y=\dfrac{t^{3}}{3}-t$ at the points where it intersects the $Ox$ axis.

Step 1: Intersections with $Ox$

\[ \frac{t^{3}}{3}-t=0 \;\Longrightarrow\; t\!\left(\frac{t^{2}}{3}-1\right)=0 \;\Longrightarrow\; t=0 \ \text{ or } \ t=\pm\sqrt3. \tag{1} \]

Points: $t=0 \Rightarrow (0,0)$; $t=\pm\sqrt3 \Rightarrow (3,0)$ — the curve passes through $(3,0)$ twice, giving two distinct tangents there.

Step 2: Parametric derivatives and general slope

\[ \dot x = 2t, \qquad \dot y = t^{2}-1, \qquad \frac{dy}{dx} = \frac{t^{2}-1}{2t}\ \ (t\neq0). \tag{2} \]

Step 3: At $t=0$, point $(0,0)$

Here $\dot x(0)=0$ but $\dot y(0)=-1\neq0$, so the tangent is vertical: $x=0$.

Step 4: At $t=\pm\sqrt3$, point $(3,0)$

\[ k\bigl(\sqrt3\bigr) = \frac{3-1}{2\sqrt3} = \frac{1}{\sqrt3}, \qquad k\bigl(-\sqrt3\bigr) = \frac{3-1}{-2\sqrt3} = -\frac{1}{\sqrt3}. \tag{3} \]

Two distinct tangents through $(3,0)$: \[y = \frac{1}{\sqrt3}(x-3) = \frac{\sqrt3}{3}(x-3), \qquad y = -\frac{1}{\sqrt3}(x-3) = -\frac{\sqrt3}{3}(x-3). \tag{4}\]

Step 5: Verification

\[t=\sqrt3\text{: } x=3,\ y=\tfrac{3\sqrt3}{3}-\sqrt3=0\ \checkmark;\qquad t=-\sqrt3\text{: same point, opposite slope}\ \checkmark\]