Parametrizing $x^2(x^2+y^2)=4(x-y)^2$

Question

I need to find $x=x(t)$ and $y=y(t)$ so that the implicitly defined curve on $\mathbb R^2$ $$x^2(x^2+y^2)=4(x-y)^2$$ is converted into an explicit function of the parameter $t$ that can be analysed using single variable calculus.

I used polar coordinates $x(\theta)=r\cos\theta$ and $y(\theta)=r\sin\theta$ and plugging it into the curve yields $$r^2\cos^2\theta\cdot r^2=-4r^2 \cdot 2\cos\theta\sin\theta,$$ which results in $$r^2(\theta)=-8\tan\theta.$$

However, plotting this curve results in a different curve than the original, see:

original: https://www.wolframalpha.com/input/?i=x%5E2(x%5E2%2By%5E2)%3D4(x-y)%5E2

polar plot: https://www.wolframalpha.com/input/?i=r%5E2%3D-8%5Ctan%5Ctheta

Can you find an error in my computation? Is it because I'm dividing by $r^2\cos^2\theta$?

EDIT: The error in my computation is found thanks to @m3801, however, the resulting function is still an implicit function. Can you please help with a different parametrization that would yield the following functions: $x=x(t)$ and $y=y(t)$?

Answer 1

HINT.-$$x^2(x^2+y^2)=4(x-y)^2\iff\left(\frac{x^2}{2(x-y)}\right)^2+\left(\frac{xy}{2(x-y)}\right)^2=1$$ Put $$\frac{x^2}{2(x-y)}=\sin (t)\space\text{ and }\space \frac{xy}{2(x-y)}=\cos (t)$$ so you can explicit after some easy algebraic calculation $$\begin{cases}x=\dfrac{2\sin (t)(1+\tan (t))}{\tan (t)}=2(\cos (t)-\sin (t))\\ \\y=2\sin (t)(1+\tan (t))\end{cases}$$

Answer 2

As you can see from the original equation, $\,x=y=r=0$ is a point on the curve. Dividing by $r^2\cos^2\theta$ is undefined.

Here's the first step after making your substitution:

$$x^2 (x^2 + y^2) = 4(x-y)^2 $$ $$\to\frac{1}{2}r^2\left(r^2 + r^2\cos(2\theta) +8\sin(2\theta) - 8\right) = 0$$

Answer 3

Given relation

$$x^2(x^2+y^2)=4(x-y)^2\tag1 $$

Divide by $x^2$ to remove singularities at origin

$$(x^2+y^2)=4(1-y/x)^2 $$

In polar coordinates

$$ r^2= 4 (1-\tan \theta)^2 $$

or $$ r = \pm 2 \sqrt{1- \tan \theta} \tag 2$$

In rectangular coordinates with $\theta$ as parameter we can have

$$ x= \pm 2 \sqrt{1- \tan \theta} \cos \theta, \quad y= \pm 2 \sqrt{1- \tan \theta} \sin \theta ; $$

An inspection of (1) shows that $x$ needs to vanish at the origin, and due to (2), that $ r $ also needs to vanish along radial line $\theta= \pi/4 $ ... as their graph also confirms.

Equation (1) can be factored. Or better, equation (2) in polar form has a $\pm$ sign for two graphs shown below:

So, no need to go in for an implicit form.