Parametric Equation of $y^2-2x^2=x^3-2y^3$

Question

Context

The context of the problem is the derivative doesn't exist at the point $(0,0)$ if you look at the curve as is in the $xy$ plane. It an intersection point. So by parametrizing the curve we can determine a suitable $t$ to determine appropriate slope given a particular $t$

Question

I'm definitely struggling with this one. The obvious $x=t$ doesn't work here as you can't write the equations as a function of $y$. I thought maybe a substitution may work but the lack of symmetry here feels like I can't actually parametrize it.

Answer 1

Using polar coordinate: $$x=\rho\cos(\theta)\qquad y=\rho\sin(\theta)$$

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$$y^{2}-2x^{2}=x^{3}-2y^{3}$$ $$\rho^{2}\sin\left(\theta\right)^{2}-2\rho^{2}\cos\left(\theta\right)^{2}=\rho^{3}\cos\left(\theta\right)^{3}-2\rho^{3}\sin\left(\theta\right)^{3}$$ $$\sin\left(\theta\right)^{2}-2\cos\left(\theta\right)^{2}=\rho\cos\left(\theta\right)^{3}-2\rho\sin\left(\theta\right)^{3}$$ $$\rho=\frac{\sin\left(\theta\right)^{2}-2\cos\left(\theta\right)^{2}}{\cos\left(\theta\right)^{3}-2\sin\left(\theta\right)^{3}}\qquad \theta\in[0,\pi]$$

If $\theta=\arctan\left(\frac{t}{2}\right)$ you have: $$\left(\frac{\sin\left(\theta\right)^{2}-2\cos\left(\theta\right)^{2}}{\cos\left(\theta\right)^{3}-2\sin\left(\theta\right)^{3}}\cdot\cos\left(\theta\right),\frac{\sin\left(t\right)^{2}-2\cos\left(t\right)^{2}}{\cos\left(t\right)^{3}-2\sin\left(t\right)^{3}}\cdot\sin\left(\theta\right)\right)$$ $$\left(\frac{8-t^{2}}{t^{3}-4},\frac{t}{2}\cdot\frac{8-t^{2}}{t^{3}-4}\right)$$

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The graph here for you

Answer 2

You can also use the substitution $y=tx$ to get a rational parametrisation:

\begin{align*} x&=\frac{t^2-2}{1-2t^3}\\ y&=\frac{(t^2-2)t}{1-2t^3} \end{align*}

Note the origin $x=y=0$ corresponds to both $t=\sqrt{2}$ and $t=-\sqrt{2}$.

One can think of this as locating the intersection of the line $y=tx$ with the curve, and since there is always a double intersection at the origin, one gets a unique `third' intersection with the cubic curve for each $t$ by Bezout's theorem

Now you get

\begin{align*} dx&=\frac{2t(1-2t^3)+6(t^2-2)t^2}{(1-2t^3)^2}dt\\ dy&=\frac{(3t^2-2)(1-2t^3)+6t(t^2-2)t^2}{(1-2t^3)^2}dt\\ \implies\frac{dy}{dx}&=\frac{(3t^2-2)(1-2t^3)+6t(t^2-2)t^2}{2t(1-2t^3)+6(t^2-2)t^2} \end{align*}