Note

This Mathematica notebook is from the WWW project A Visual Dictionary of Special Plane Curves. Copyright © 2004 by Xah Lee. (http://xahlee.org/)

Proof of different formulas for astroid

Problem

Given the parametric formula
x=Cos[t]^3
y=Sin[t]^3
find the Cartesian equation for it.

Solution

Converting the Parametric formula to Cartesian equation

If x == Cos[t]^3, y == Sin[t]^3, then raise both side to the power of 2/3 we have
(x^(2/3)) == Cos[t]^2
(y^(2/3)) == Sin[t]^2
(We do not have to worry about the sign in squaring both sides, because x==Cos[t]^3 is a definition, not an equation with an unknown.)
Adding these equations results in
(x^(2/3)) + (y^(2/3)) == (Cos[t]^2 + Sin[t]^2)
(x^(2/3)) + (y^(2/3)) == 1

Expanding the Equation

(x^(2/3)) + (y^(2/3)) == 1, from this we can get several equivalent forms, to be used later. ( (*3*) is not strickly equivalent because we applied the non-one-to-one function #^(2) to both sides)

(x^(2/3)) == 1 - (y^(2/3))	(*1*)
(y^(2/3)) == 1 - (x^(2/3))	(*2*)
1 == (x^(4/3)) + (y^(4/3)) + 2 (x^(2/3)) (y^(2/3))	(*3*)

(x^(2/3)) + (y^(2/3)) == 1 apply the function #^3& to both sides
(x^2) + (y^2) + 3 (x^(4/3)) (y^(2/3)) + 
          3 (x^(2/3)) (y^(4/3)) == 1
3 ((x^(4/3)) (y^(2/3)) + (x^(2/3)) (y^(4/3))) == 
        1 - (x^2) - (y^2)

substitute (x^(2/3)) and (y^(2/3)) using (*1*) and (*2*),
substitute (x^(4/3)) + (y^(4/3)) using (*3*)
simplify and apply the function #^3 one more time to get
((((x^2) + (y^2) - 1))^3) + 27 (x^3) (y^3) == 0.

here we do it in Mathematica.

ast = {x^(2/3) + y^(2/3), 1}

ast1 = ast - y^(2/3) // Expand

ast2 = ast - x^(2/3) // Expand

ast3 = ast^2 - 2  x^(2/3)*y^(2/3) // Expand

ast4 = (ast^3 - x^2 - y^2)/3 // Expand

ast4 = ast4*3 // Together

ast5 = ast4 /. {Rule @@ ast1, Rule @@ ast2}

ast6 = (ast5 + 2*x^2 + 2*y^2) // Expand // Factor

ast7 = ast6 /. Rule @@ ast3

ast8 = -Expand@ast7 + 3

ast9 = ast8^3

Now we have an equation without fraction powers. Astroid is commonly given as
(x^2+y^2-a^2)^3+27 a^2 x^2 y^2==0
which is derived by starting with the parametric formula with a scaling factor a. Here we show how to arrive at that form from our equation above.

ast9/8

ast10 = ast9/8 // Expand // Factor

Now, we introduce a scaling factor by replacing x with x/a and y with y/a.

ast11 = ast10 /. {x -> x/a, y -> y/a}

ast11 // Expand // Factor

%*a^6