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Cartesian? Check me please?
x^6+3 x^4 (y^2-3)+3 x^2 y^2 (y^2+2)+y^6 = y^4
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\[x^6+3 x^4 (y^2-3)+3 x^2 y^2 (y^2+2)+y^6 = y^4\]

I just chose some arbitrary "a" constant to sub.

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\(x^2 + y^2 = \text{radius}^2\)

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so a=1

\(\text{Area} = 3 \pi a^2\), correct ?

Does this graph actually have a name though? It's not as if I can just Google the image lol

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*reading & taking notes* :-3

origin build with 1+cos2t=0 _> x=0 , y=0 common term between x , y

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is it clear?

for exactly drawing d/dt=0 for x , y give you some critical points

Clear, you are finding the polar lines, basically, where r = 0.

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dy/dt = 0 is for horizontal I believe

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interesting problem !

Does this funky thing have a name though? :-D

No! it has a nice shape .

It kind of looks like... one of those things when you roll a circle within a circle

Try to solve one by yourself then think it's so easy and sure after that enjoy with the figure.

Well I'm not going to sweat it, the hard part is done. Who cares about the name, you're right :-D

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first thing attracted me to math was figuring out the shape of equation.

dx = 3 sin(3t) dt
dy = 2 cos(t)-3 cos(3t) dt

Are you solving mine?

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