Tangents with Parametric Equations — Question 5

Question 5 – Solution

Solution

1. At a self-intersection, two distinct parameter values give the same $y$-coordinate. Since $y=t^2$, equal $y$-coordinates occur for parameters $t$ and $-t$.

2. Require the corresponding $x$-coordinates to be equal: $$t^3-3t=(-t)^3-3(-t)=-t^3+3t.$$ Simplifying gives $$2t^3-6t=2t(t^2-3)=0.$$ The value $t=0$ does not give two distinct parameters. The nontrivial pair is $$t=\pm\sqrt3.$$

3. At either parameter value, $$x=t(t^2-3)=0, \qquad y=t^2=3.$$ Hence the self-intersection is $\boxed{(0,3)}$.

4. Differentiate: $$\frac{dx}{dt}=3t^2-3, \qquad \frac{dy}{dt}=2t.$$ At $t=\pm\sqrt3$, $dx/dt=6\ne0$, so $$\frac{dy}{dx}=\frac{2t}{3t^2-3}=\frac{t}{3}.$$ The two slopes are $$m_1=\frac1{\sqrt3}, \qquad m_2=-\frac1{\sqrt3}.$$

5. Two nonvertical lines are perpendicular only if the product of their slopes is $-1$. Here $$m_1m_2=-\frac13\ne-1.$$ Therefore, the two tangents are .