Answer
(a)
The parametrization of the folium of Descartes is
$\left(\frac{3 a t}{1+t^3},\frac{3 a t^2}{1+t^3}\right)$, excluding $t\neq -1,0$.
The arrows indicate the direction of motion.
(b)
Quadrant I: $t>0$
Quadrant II: $-1
Work Step by Step
(a)
We find the intersection of $y=t x$ and the curve by substituting $y=t x$ into $x^3+y^3=3a x y$, which gives
$x^3+t^3 x^3=3a t x^2$,
$x^3(1+t^3)-3a t x^2=0$,
$x^2 (x (1+t^3)-3a t)=0$,
The solutions are $x=0$ or $x=\frac{3 a t}{1+t^3}$. So, we have $y=\frac{3 a t^2}{1+t^3}$.
If $t=0$, this is the origin, where $x=y=0$. If $t=-1$, $x$ becomes infinite. So, we exclude $t=-1$. Thus, the line $y=t x$ intersects the folium at the origin and at one other point P for all $t\neq -1,0$. The coordinates of P is $P=\left(\frac{3 a t}{1+t^3},\frac{3 a t^2}{1+t^3}\right)$. Since $t$ is arbitrary, these coordinates determine all points on the folium of Descartes except at $t\neq -1,0$. We conclude that the folium of Descartes has parametrization $\left(\frac{3 a t}{1+t^3},\frac{3 a t^2}{1+t^3}\right)$, excluding $t\neq -1,0$.
To find the direction of motion, we take $a=1$ and plot the curve for several $t$-values as is shown in the figure attached. The arrows indicate the direction of motion.
(b)
Let us consider $a>0$.
Quadrant I: $x>0$ and $y>0$
The part of the curve is in quadrant I if $x>0$ and $y>0$, that is,
$\frac{3 a t}{1+t^3}>0$ and $\frac{3 a t^2}{1+t^3}>0$.
Since $\frac{3 a t}{1+t^3}>0$, we must have $t>0$ such that $\frac{3 a t^2}{1+t^3}>0$. So the interval of $t$-values is $t>0$.
Quadrant II: $x<0$ and $y>0$
The part of the curve is in quadrant II if $x<0$ and $y>0$, that is,
$\frac{3 a t}{1+t^3}<0$ and $\frac{3 a t^2}{1+t^3}>0$.
Since $\frac{3 a t}{1+t^3}<0$, we must have $t<0$ such that $\frac{3 a t^2}{1+t^3}>0$. But $\frac{3 a t}{1+t^3}<0$, so $1+t^2>0$. This means that $-10$ and $y<0$
The part of the curve is in quadrant IV if $x>0$ and $y<0$, that is,
$\frac{3 a t}{1+t^3}>0$ and $\frac{3 a t^2}{1+t^3}<0$.
Since $\frac{3 a t}{1+t^3}>0$, we must have $t<0$ such that $\frac{3 a t^2}{1+t^3}<0$. But $\frac{3 a t}{1+t^3}>0$, so $1+t^3<0$. This means that $t
