The tangent to the curve $y=x-x^3$ at point P meets the curve again at point Q.

Question

One point of trisection lies on $y$ axis. What is the equation of locus of other point of trisection?

I have tried finding the slope, distance and tangent equation but ended up with no results and a tedious work. Can someone help me in giving an insight to this problem and its solution?

Answer 1

Let $P(p,p-p^3)$. Since $y'=1-3x^2$, the equation of the tangent line at $P$ is given by $$y-(p-p^3)=(1-3p^2)(x-p)$$ We want to find $x\not=p$ such that $$(x-x^3)-(p-p^3)=(1-3p^2)(x-p),$$ i.e. $$x^3-3p^2x+2p^3=0$$ Since the LHS is divisible by $(x-p)^2$, we obtain $$(x-p)^2(x+2p)=0.$$ Hence, $Q(-2p, 8p^3-2p)$. I think that you can continue from here.

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For the trisection points $R_1,R_2$ of the line segment $PQ$ :

$R_1(x_1,y_1)$ : $$x_1=\frac{2\times p+1\times (-2p)}{3}=0,\quad y_1=\frac{2\times (p-p^3)+1\times (8p^3-2p)}{3}$$

$R_2(x_2,y_2)$ : $$x_2=\frac{1\times p+2\times (-2p)}{3},\quad y_2=\frac{1\times (p-p^3)+2\times (8p^3-2p)}{3}$$

Answer 2

Given $y=x-x^3$ and a tangent at point $P(a,a-a^3)$ also intersects the graph at a point $Q$.

The slope of the tangent line is $y=1-3x^2$, so the slope of the tangent line at $(a,a-a^3)$ on the graph is $m=1-3a^2$.

So the equation of the tangent line at that point is

\begin{equation} y-a+a^3=(1-3a^2)(x-a) \end{equation} or \begin{equation} y=(1-3a^2)x+2a^3 \end{equation}

Since $Q$ lies on the tangent line and also on the curve, the $x$ coordinate of $Q$ satisfies the equation

\begin{equation} x-x^3=(1-3a^2)x+2a^3 \end{equation}

which can be re-written

\begin{equation} x^3-3a^2x+2a^3=0 \end{equation}

Since $P$ also lies on the tangent line and the curve we know that $x=a$ is one solution of this equation. Thus we may factor out $x-a$ to obtain the equation

\begin{equation} (x-a)(x^2+ax-2a^2)=0 \end{equation}

and the quadratic can also be factored to give

\begin{equation} (x-a)^2(x+2a)=0 \end{equation}

We should not be surprised to see that $a$ is a double solution since that is where the tangent it. But the tangent line crosses the graph at $x=-2a$ so that is a single root of the polynomial.

The coordinates of $Q$ are $(-2a,-2a+8a^3)$.