How to find point in Descartes Folium with slope of -1/3?

Question

I stumbled upon a problem in my calculus book that asked to find the point in $x^3 + y^3 = 3xy$ that had a slope perpendicualr to $y = 3x + 1$ and also was in the first quadrant.

I began by getting the derivative of the equation and got $\dfrac{x^2-y}{x-y^2}$.

Given that the slope is $-\frac13$, I wrote $\dfrac{x^2-y}{x-y^2}=-\frac13 $, which got me to $3x^2+x=y^2+3y$, and that is where I got stuck, because I cannot solve for $x$ or $y$, to substitute in the original equation.

I looked for similar examples on the internet, and only found some where the slope was $0$, which can be solved because $y = x^2$, which can be substituted back.

Hope this is redacted well enough, any help is really appreciated.

Answer 1

Let's find a parametric equation of the curve defined by $x^3+y^3=3xy$.

Let $y=xt$, then

$$x^3(1+t^3)=3x^2t$$

Hence

$$x=\frac{3t}{1+t^3}$$ $$y=\frac{3t^2}{1+t^3}$$

Now, the tangent vector is given by

$$x'(t)=\frac{3(1+t^3)-3t(3t^2)}{(1+t^3)^2}=\frac{3-6t^3}{(1+t^3)^2}$$ $$y'(t)=\frac{6t(1+t^3)-3t^2(3t^2)}{(1+t^3)^2}=\frac{6t-3t^4}{(1+t^3)^2}$$

You want $t$ such that $y'(t)=-\frac13x'(t)$, that is

$$6t-3t^4=2t^3-1$$

You have thus a quartic equation to solve: $3t^4+2t^3-6t-1=0$. There does not seem to be a trivial solution, so you'll have to go through Descartes' or Ferrari's method to solve this, or a numerical method.

There are two real roots (hence two points), with numerically

$$t_1=-0.1678460300438111, x_1=-0.5059304363344097, y_1=0.0849184152170638$$ $$t_2=1.1301467382482941, x_2=1.3875575405757694, y_2=1.5681436286135308$$

Answer 2

I cannot understand what you mean "had a slope perpendicular to $y=3x+1$ ".

A slope is an number and $y=3x+1$ is the equation of a straight line. A number perpendicular to a line is non-sens.

I suppose that you mean "to find the point where the tangent to $x^3 + y^3 = 3xy$ is perpendicular to $y = 3x + 1$ "

If so, the slope of the tangent must be $-\frac13$.

Differentiating $x^3 + y^3 = 3xy$ leads to : $$3x^2dx+3y^2dy=3ydx+3xdy$$ $$\frac{dy}{dx}=\frac{y-x^2}{y^2-x}$$ All the points $(x,y)$ where the slope of the tangent is $-\frac13$ are given by the system of equations : $$\begin{cases} x^3 + y^3 = 3xy \\ \frac{y-x^2}{y^2-x}=-\frac13 \end{cases}$$ Solve the system of equations for the points $(x,y)$ and select the point in the first quadrant. (Numerical solving).

In addition. The analytical solving leads to the exact formulas :