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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 $y=y(t)$?

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    $\begingroup$ If you try to convert to polar coordinates, I guess the error is in $x-y = r(\cos \theta - \sin \theta)$, which gives $4(x-y)^2 = 4r^2-4r^2\sin 2\theta$. $\endgroup$
    – Gibbs
    Commented Mar 21, 2018 at 18:28

3 Answers 3

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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 $$\begin{cases}x=\dfrac{2\sin (t)(1+\tan (t))}{\tan (t)}=2(\cos (t)-\sin (t))\\ \\y=2\sin (t)(1+\tan (t))\end{cases}$$

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  • $\begingroup$ The substitution works great, only the algebraic calculation leads to $x=2(\sin(t)-\cos(t))$ and $y=2\cos(t)(1-\cot(t))$, not the one you mentioned. However, your answer helped the most, thank you. $\endgroup$
    – mathearts
    Commented Mar 21, 2018 at 21:45
  • $\begingroup$ It is true. Thanks. $\endgroup$
    – Piquito
    Commented Mar 21, 2018 at 22:19
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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:

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

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  • $\begingroup$ The resulting function is still an implicit function. Would a different parametrization help? $\endgroup$
    – mathearts
    Commented Mar 21, 2018 at 20:42
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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:

enter image description here

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

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  • $\begingroup$ DIviding by $x^2$ is a great trick! I'll go with Piquito's answer because that parametrization yields no square roots and is thus easier to work with, but your solution is extremely elegant, I appreciate it. $\endgroup$
    – mathearts
    Commented Mar 21, 2018 at 22:12
  • $\begingroup$ Actually, this red curve is different to the one in the original question (see wolfram link above). This transformation is thus not regular but I can't seem to find a possible misstep in your calculation except dividing by $x^2$... $\endgroup$
    – mathearts
    Commented Mar 22, 2018 at 16:41
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    $\begingroup$ The two curves (red,blue) now added are quite regular and well behaved with continuous derivatives, no misstep as shown. They represent the entire double equation (1). It is like two straight lines that are factorable for degenate conics. $\endgroup$
    – Narasimham
    Commented Mar 22, 2018 at 17:57
  • $\begingroup$ @ mathearts No try plotting them separately. $\endgroup$
    – Narasimham
    Commented Mar 25, 2018 at 16:34

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