For the curve x=t², y=t³, find the pairs of points
For the curve $x=t^{2},\ y=t^{3}$, find the pairs of points at which the tangent line at one point coincides with the normal line at the other — and show that this happens at the parameter values $t=\pm\dfrac{\sqrt2}{3}$ (together with their partners).
Explanation of Method
One tangent line, playing the role of tangent at one point and normal at another.
Step 1: Slope along the curve
\[ \dot x = 2t, \qquad \dot y = 3t^{2}, \qquad \frac{dy}{dx} = \frac{3t}{2}\ \ (t\neq0). \tag{1} \]
Step 2: Where the tangent at $t_{0}$ meets the curve again
The tangent at $t_{0}$ is $y-t_{0}^{3}=\dfrac{3t_{0}}{2}(x-t_{0}^{2})$. Requiring the point $(t_{1}^{2},t_{1}^{3})$ to lie on it, and dividing by the common factor $(t_{1}-t_{0})$:
\[ t_{1}^{2}+t_{1}t_{0}+t_{0}^{2} = \frac{3t_{0}}{2}(t_{1}+t_{0}) \;\Longrightarrow\; (t_{1}-t_{0})(2t_{1}+t_{0})=0, \]
so besides $t_{1}=t_{0}$ itself, the tangent at $t_{0}$ meets the curve again at
\[ t_{1} = -\frac{t_{0}}{2}. \tag{2} \]
(This is the same double-root mechanism used in Problems 505–507 above, now for the semicubical parabola $y^{2}=x^{3}$.)
Step 3: When is that same tangent also the normal at $t_{1}$?
The tangent at $t_{0}$ already passes through the point at $t_{1}=-t_{0}/2$ by construction; it coincides with the normal at $t_{1}$ exactly when their slopes agree, i.e.\ when the tangent slope at $t_{0}$ equals $-1/(\text{tangent slope at } t_{1})$:
\[ \frac{3t_{0}}{2} = -\frac{1}{\dfrac{3t_{1}}{2}} = -\frac{2}{3t_{1}}. \tag{3} \]
Substituting $t_{1}=-t_{0}/2$: \[\frac{3t_{0}}{2} = -\frac{2}{3\left(-\dfrac{t_{0}}{2}\right)} = \frac{4}{3t_{0}} \;\Longrightarrow\; 9t_{0}^{2}=8 \;\Longrightarrow\; t_{0} = \pm\frac{2\sqrt2}{3}. \tag{4}\]
Step 4: The paired parameter
From $t_{1}=-t_{0}/2$: \[t_{0}=\frac{2\sqrt2}{3} \;\Longrightarrow\; t_{1}=-\frac{\sqrt2}{3}, \qquad\qquad t_{0}=-\frac{2\sqrt2}{3} \;\Longrightarrow\; t_{1}=\frac{\sqrt2}{3}. \tag{5}\]
Step 5: Verification
\[t_1=\sqrt2/3,\ t_0=-2\sqrt2/3\text{: tangent slope at } t_0 \text{ equals normal slope at } t_1 \text{ (both} \approx-1.41421)\ \checkmark\]
\[\text{Tangent line at } t_0 \text{ passes through the point at } t_1\text{, and vice versa (numerically, residual} \sim10^{-16})\ \checkmark\]
Result
Final answer:
\[\boxed{\,\left(t_{0},t_{1}\right) = \left(\frac{2\sqrt2}{3},\,-\frac{\sqrt2}{3}\right) \quad\text{and}\quad \left(-\frac{2\sqrt2}{3},\,\frac{\sqrt2}{3}\right)\,}\]
At these pairs, the tangent to the curve at $t_{0}$ is exactly the normal to the curve at $t_{1}=\pm\sqrt2/3$, and conversely. $\blacksquare$
Remark
Remark. This is the cleanest way to see the claim: rather than solving simultaneously for “tangent through the other point” and “perpendicular slopes” as two unrelated equations, note that the first condition is already forced by the geometry of a cubic (double root at $t_0$, third root at $-t_0/2$), so only the slope condition (3) remains free — and it is a single quadratic in $t_0$.