Express a parametrization of a curve $C$ as a function of $X$ and $Y$












3












$begingroup$


Express the curve $C$ with parametrization ${(cos 3t, sin 2t) : t in mathbb{R} }$ as a function of $X$ and $Y$.



Here's what I have (I don't know if this is helpful)



I let $X= cos 3t$ and $Y= sin 2t$. So I have $X= cos^3 t -3sin^2 t cos t$ and $Y=2 cos t sin t$.



I don't know how to go on from here. Please help. Thanks!






algebraic-geometry







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$endgroup$












  • $begingroup$
    As a set of points, $C$ isn't a function from $mathbb R$ tp $mathbb R$.
    $endgroup$
    – Michael Hoppe
    Dec 3 '18 at 16:25
















3












$begingroup$


Express the curve $C$ with parametrization ${(cos 3t, sin 2t) : t in mathbb{R} }$ as a function of $X$ and $Y$.



Here's what I have (I don't know if this is helpful)



I let $X= cos 3t$ and $Y= sin 2t$. So I have $X= cos^3 t -3sin^2 t cos t$ and $Y=2 cos t sin t$.



I don't know how to go on from here. Please help. Thanks!






algebraic-geometry







share|cite|improve this question









$endgroup$












  • $begingroup$
    As a set of points, $C$ isn't a function from $mathbb R$ tp $mathbb R$.
    $endgroup$
    – Michael Hoppe
    Dec 3 '18 at 16:25














3












3








3





$begingroup$


Express the curve $C$ with parametrization ${(cos 3t, sin 2t) : t in mathbb{R} }$ as a function of $X$ and $Y$.



Here's what I have (I don't know if this is helpful)



I let $X= cos 3t$ and $Y= sin 2t$. So I have $X= cos^3 t -3sin^2 t cos t$ and $Y=2 cos t sin t$.



I don't know how to go on from here. Please help. Thanks!






algebraic-geometry







share|cite|improve this question









$endgroup$




Express the curve $C$ with parametrization ${(cos 3t, sin 2t) : t in mathbb{R} }$ as a function of $X$ and $Y$.



Here's what I have (I don't know if this is helpful)



I let $X= cos 3t$ and $Y= sin 2t$. So I have $X= cos^3 t -3sin^2 t cos t$ and $Y=2 cos t sin t$.



I don't know how to go on from here. Please help. Thanks!






algebraic-geometry




algebraic-geometry






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 11:39









Mashed PotatoMashed Potato

866




866












  • $begingroup$
    As a set of points, $C$ isn't a function from $mathbb R$ tp $mathbb R$.
    $endgroup$
    – Michael Hoppe
    Dec 3 '18 at 16:25


















  • $begingroup$
    As a set of points, $C$ isn't a function from $mathbb R$ tp $mathbb R$.
    $endgroup$
    – Michael Hoppe
    Dec 3 '18 at 16:25
















$begingroup$
As a set of points, $C$ isn't a function from $mathbb R$ tp $mathbb R$.
$endgroup$
– Michael Hoppe
Dec 3 '18 at 16:25




$begingroup$
As a set of points, $C$ isn't a function from $mathbb R$ tp $mathbb R$.
$endgroup$
– Michael Hoppe
Dec 3 '18 at 16:25










1 Answer
1






active

oldest

votes


















2












$begingroup$

Answer:
$$
X^4 - X^2 + 4 Y^6 - 6 Y^4 + frac{9}{4} Y^2 = 0
$$



How I got it:



We can replace the trigonometric parametrization $(cos(u), sin(u))$ of the unit circle with a rational one: $left(frac{1-t^2}{1+t^2}, frac{2t}{1+t^2}right)$. So we equate $cos(u) = x = frac{1-t^2}{1+t^2}$ and $sin(u) = y =frac{1-t^2}{1+t^2}$. Your expressions for $X$ and $Y$ then become
begin{align*}
X &= cos^3(u) - 3 cos(u) sin^2(u) = x^3 - 3 xy^2 = frac{-t^{6} + 15 t^{4} - 15 t^{2} + 1}{(1+t^2)^3}\
Y &= 2 cos(u) sin(u) = 2 x y = frac{-4 t^{3} + 4 t}{(1+t^2)^2}
end{align*}



In the polynomial ring $mathbb{Q}[t,z,X,Y]$ we form the ideal
$$
I = ((1+t^2)^3 X -(-t^{6} + 15 t^{4} - 15 t^{2} + 1), (1+t^2)^2 Y - (-4 t^{3} + 4 t), (1+t^2)z - 1)
$$
and compute a Gröbner basis with respect to the elimination ordering $t > z > X > Y$. The last entry of this Gröbner basis is the desired polynomial in $X$ and $Y$.



The proof is in the pudding, though: here is a SageMathCell comparing the two plots.



Actually, a slightly better way is to instead compute a Gröbner basis for the ideal $(X - (x^3 - 3xy^2), Y - (2xy), x^2 + y^2 - 1)$ in $mathbb{Q}[x,y,X,Y]$ with respect to the elimination ordering $x > y > X > Y$. This yields the same answer.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Another way to get rid of fractions is to homogenize to I=ideal(X-(-t^6+15*t^4*u^2-15*t^2*u^4+u^6),Y-(-4*t^3*u+4*t*u^3)*(t^2+u^2),Z-(t^2+u^2)^3) and after taking the gröbner basis, set $Z=1$ in the resulting $16Y^6+4X^4Z^2-24Y^4Z^2-4X^2Z^4+9Y^2Z^4$. (+1)
    $endgroup$
    – Jan-Magnus Økland
    Dec 3 '18 at 18:21










  • $begingroup$
    Thanks for the detailed reply! However, we haven't encounterer Gröbner basis yet. Do you think there is still another way to get it without using Gröbner basis?
    $endgroup$
    – Mashed Potato
    Dec 3 '18 at 22:05










  • $begingroup$
    @MashedPotato Have you learned other tools from elimination theory? It is probably possible to do this with resultants instead of Gröbner bases. A more brute-force approach might also be possible. From a plot of the parametrization, we can see that the curve has degree $6$ (a generic line intersects it in $6$ points). So we write down a degree $6$ polynomial with $28$ undetermined coefficients. We can impose linear conditions conditions on these coefficients by plugging in points we obtain by substituting various values of $t$ into the parametrization, e.g., $t=0$ says $(1,0)$ lies on $C$.
    $endgroup$
    – André 3000
    Dec 3 '18 at 23:00










  • $begingroup$
    You can also plug the parametrization directly into the degree $6$ polynomial in $X$ and $Y$ and get linear relations that way. But this naïve method seems quite hard and tedious! The thing is, once you have a proposed answer, it's easy to check by substituting in the parametrization and seeing that the equation holds.
    $endgroup$
    – André 3000
    Dec 3 '18 at 23:00








  • 1




    $begingroup$
    I'm only taking an introductory course so I think what will really work for me are the brute force methods. Thanks for your help!!
    $endgroup$
    – Mashed Potato
    Dec 4 '18 at 2:59











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Answer:
$$
X^4 - X^2 + 4 Y^6 - 6 Y^4 + frac{9}{4} Y^2 = 0
$$



How I got it:



We can replace the trigonometric parametrization $(cos(u), sin(u))$ of the unit circle with a rational one: $left(frac{1-t^2}{1+t^2}, frac{2t}{1+t^2}right)$. So we equate $cos(u) = x = frac{1-t^2}{1+t^2}$ and $sin(u) = y =frac{1-t^2}{1+t^2}$. Your expressions for $X$ and $Y$ then become
begin{align*}
X &= cos^3(u) - 3 cos(u) sin^2(u) = x^3 - 3 xy^2 = frac{-t^{6} + 15 t^{4} - 15 t^{2} + 1}{(1+t^2)^3}\
Y &= 2 cos(u) sin(u) = 2 x y = frac{-4 t^{3} + 4 t}{(1+t^2)^2}
end{align*}



In the polynomial ring $mathbb{Q}[t,z,X,Y]$ we form the ideal
$$
I = ((1+t^2)^3 X -(-t^{6} + 15 t^{4} - 15 t^{2} + 1), (1+t^2)^2 Y - (-4 t^{3} + 4 t), (1+t^2)z - 1)
$$
and compute a Gröbner basis with respect to the elimination ordering $t > z > X > Y$. The last entry of this Gröbner basis is the desired polynomial in $X$ and $Y$.



The proof is in the pudding, though: here is a SageMathCell comparing the two plots.



Actually, a slightly better way is to instead compute a Gröbner basis for the ideal $(X - (x^3 - 3xy^2), Y - (2xy), x^2 + y^2 - 1)$ in $mathbb{Q}[x,y,X,Y]$ with respect to the elimination ordering $x > y > X > Y$. This yields the same answer.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Another way to get rid of fractions is to homogenize to I=ideal(X-(-t^6+15*t^4*u^2-15*t^2*u^4+u^6),Y-(-4*t^3*u+4*t*u^3)*(t^2+u^2),Z-(t^2+u^2)^3) and after taking the gröbner basis, set $Z=1$ in the resulting $16Y^6+4X^4Z^2-24Y^4Z^2-4X^2Z^4+9Y^2Z^4$. (+1)
    $endgroup$
    – Jan-Magnus Økland
    Dec 3 '18 at 18:21










  • $begingroup$
    Thanks for the detailed reply! However, we haven't encounterer Gröbner basis yet. Do you think there is still another way to get it without using Gröbner basis?
    $endgroup$
    – Mashed Potato
    Dec 3 '18 at 22:05










  • $begingroup$
    @MashedPotato Have you learned other tools from elimination theory? It is probably possible to do this with resultants instead of Gröbner bases. A more brute-force approach might also be possible. From a plot of the parametrization, we can see that the curve has degree $6$ (a generic line intersects it in $6$ points). So we write down a degree $6$ polynomial with $28$ undetermined coefficients. We can impose linear conditions conditions on these coefficients by plugging in points we obtain by substituting various values of $t$ into the parametrization, e.g., $t=0$ says $(1,0)$ lies on $C$.
    $endgroup$
    – André 3000
    Dec 3 '18 at 23:00










  • $begingroup$
    You can also plug the parametrization directly into the degree $6$ polynomial in $X$ and $Y$ and get linear relations that way. But this naïve method seems quite hard and tedious! The thing is, once you have a proposed answer, it's easy to check by substituting in the parametrization and seeing that the equation holds.
    $endgroup$
    – André 3000
    Dec 3 '18 at 23:00








  • 1




    $begingroup$
    I'm only taking an introductory course so I think what will really work for me are the brute force methods. Thanks for your help!!
    $endgroup$
    – Mashed Potato
    Dec 4 '18 at 2:59
















2












$begingroup$

Answer:
$$
X^4 - X^2 + 4 Y^6 - 6 Y^4 + frac{9}{4} Y^2 = 0
$$



How I got it:



We can replace the trigonometric parametrization $(cos(u), sin(u))$ of the unit circle with a rational one: $left(frac{1-t^2}{1+t^2}, frac{2t}{1+t^2}right)$. So we equate $cos(u) = x = frac{1-t^2}{1+t^2}$ and $sin(u) = y =frac{1-t^2}{1+t^2}$. Your expressions for $X$ and $Y$ then become
begin{align*}
X &= cos^3(u) - 3 cos(u) sin^2(u) = x^3 - 3 xy^2 = frac{-t^{6} + 15 t^{4} - 15 t^{2} + 1}{(1+t^2)^3}\
Y &= 2 cos(u) sin(u) = 2 x y = frac{-4 t^{3} + 4 t}{(1+t^2)^2}
end{align*}



In the polynomial ring $mathbb{Q}[t,z,X,Y]$ we form the ideal
$$
I = ((1+t^2)^3 X -(-t^{6} + 15 t^{4} - 15 t^{2} + 1), (1+t^2)^2 Y - (-4 t^{3} + 4 t), (1+t^2)z - 1)
$$
and compute a Gröbner basis with respect to the elimination ordering $t > z > X > Y$. The last entry of this Gröbner basis is the desired polynomial in $X$ and $Y$.



The proof is in the pudding, though: here is a SageMathCell comparing the two plots.



Actually, a slightly better way is to instead compute a Gröbner basis for the ideal $(X - (x^3 - 3xy^2), Y - (2xy), x^2 + y^2 - 1)$ in $mathbb{Q}[x,y,X,Y]$ with respect to the elimination ordering $x > y > X > Y$. This yields the same answer.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Another way to get rid of fractions is to homogenize to I=ideal(X-(-t^6+15*t^4*u^2-15*t^2*u^4+u^6),Y-(-4*t^3*u+4*t*u^3)*(t^2+u^2),Z-(t^2+u^2)^3) and after taking the gröbner basis, set $Z=1$ in the resulting $16Y^6+4X^4Z^2-24Y^4Z^2-4X^2Z^4+9Y^2Z^4$. (+1)
    $endgroup$
    – Jan-Magnus Økland
    Dec 3 '18 at 18:21










  • $begingroup$
    Thanks for the detailed reply! However, we haven't encounterer Gröbner basis yet. Do you think there is still another way to get it without using Gröbner basis?
    $endgroup$
    – Mashed Potato
    Dec 3 '18 at 22:05










  • $begingroup$
    @MashedPotato Have you learned other tools from elimination theory? It is probably possible to do this with resultants instead of Gröbner bases. A more brute-force approach might also be possible. From a plot of the parametrization, we can see that the curve has degree $6$ (a generic line intersects it in $6$ points). So we write down a degree $6$ polynomial with $28$ undetermined coefficients. We can impose linear conditions conditions on these coefficients by plugging in points we obtain by substituting various values of $t$ into the parametrization, e.g., $t=0$ says $(1,0)$ lies on $C$.
    $endgroup$
    – André 3000
    Dec 3 '18 at 23:00










  • $begingroup$
    You can also plug the parametrization directly into the degree $6$ polynomial in $X$ and $Y$ and get linear relations that way. But this naïve method seems quite hard and tedious! The thing is, once you have a proposed answer, it's easy to check by substituting in the parametrization and seeing that the equation holds.
    $endgroup$
    – André 3000
    Dec 3 '18 at 23:00








  • 1




    $begingroup$
    I'm only taking an introductory course so I think what will really work for me are the brute force methods. Thanks for your help!!
    $endgroup$
    – Mashed Potato
    Dec 4 '18 at 2:59














2












2








2





$begingroup$

Answer:
$$
X^4 - X^2 + 4 Y^6 - 6 Y^4 + frac{9}{4} Y^2 = 0
$$



How I got it:



We can replace the trigonometric parametrization $(cos(u), sin(u))$ of the unit circle with a rational one: $left(frac{1-t^2}{1+t^2}, frac{2t}{1+t^2}right)$. So we equate $cos(u) = x = frac{1-t^2}{1+t^2}$ and $sin(u) = y =frac{1-t^2}{1+t^2}$. Your expressions for $X$ and $Y$ then become
begin{align*}
X &= cos^3(u) - 3 cos(u) sin^2(u) = x^3 - 3 xy^2 = frac{-t^{6} + 15 t^{4} - 15 t^{2} + 1}{(1+t^2)^3}\
Y &= 2 cos(u) sin(u) = 2 x y = frac{-4 t^{3} + 4 t}{(1+t^2)^2}
end{align*}



In the polynomial ring $mathbb{Q}[t,z,X,Y]$ we form the ideal
$$
I = ((1+t^2)^3 X -(-t^{6} + 15 t^{4} - 15 t^{2} + 1), (1+t^2)^2 Y - (-4 t^{3} + 4 t), (1+t^2)z - 1)
$$
and compute a Gröbner basis with respect to the elimination ordering $t > z > X > Y$. The last entry of this Gröbner basis is the desired polynomial in $X$ and $Y$.



The proof is in the pudding, though: here is a SageMathCell comparing the two plots.



Actually, a slightly better way is to instead compute a Gröbner basis for the ideal $(X - (x^3 - 3xy^2), Y - (2xy), x^2 + y^2 - 1)$ in $mathbb{Q}[x,y,X,Y]$ with respect to the elimination ordering $x > y > X > Y$. This yields the same answer.






share|cite|improve this answer











$endgroup$



Answer:
$$
X^4 - X^2 + 4 Y^6 - 6 Y^4 + frac{9}{4} Y^2 = 0
$$



How I got it:



We can replace the trigonometric parametrization $(cos(u), sin(u))$ of the unit circle with a rational one: $left(frac{1-t^2}{1+t^2}, frac{2t}{1+t^2}right)$. So we equate $cos(u) = x = frac{1-t^2}{1+t^2}$ and $sin(u) = y =frac{1-t^2}{1+t^2}$. Your expressions for $X$ and $Y$ then become
begin{align*}
X &= cos^3(u) - 3 cos(u) sin^2(u) = x^3 - 3 xy^2 = frac{-t^{6} + 15 t^{4} - 15 t^{2} + 1}{(1+t^2)^3}\
Y &= 2 cos(u) sin(u) = 2 x y = frac{-4 t^{3} + 4 t}{(1+t^2)^2}
end{align*}



In the polynomial ring $mathbb{Q}[t,z,X,Y]$ we form the ideal
$$
I = ((1+t^2)^3 X -(-t^{6} + 15 t^{4} - 15 t^{2} + 1), (1+t^2)^2 Y - (-4 t^{3} + 4 t), (1+t^2)z - 1)
$$
and compute a Gröbner basis with respect to the elimination ordering $t > z > X > Y$. The last entry of this Gröbner basis is the desired polynomial in $X$ and $Y$.



The proof is in the pudding, though: here is a SageMathCell comparing the two plots.



Actually, a slightly better way is to instead compute a Gröbner basis for the ideal $(X - (x^3 - 3xy^2), Y - (2xy), x^2 + y^2 - 1)$ in $mathbb{Q}[x,y,X,Y]$ with respect to the elimination ordering $x > y > X > Y$. This yields the same answer.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 17:28

























answered Dec 3 '18 at 17:07









André 3000André 3000

12.5k22143




12.5k22143








  • 1




    $begingroup$
    Another way to get rid of fractions is to homogenize to I=ideal(X-(-t^6+15*t^4*u^2-15*t^2*u^4+u^6),Y-(-4*t^3*u+4*t*u^3)*(t^2+u^2),Z-(t^2+u^2)^3) and after taking the gröbner basis, set $Z=1$ in the resulting $16Y^6+4X^4Z^2-24Y^4Z^2-4X^2Z^4+9Y^2Z^4$. (+1)
    $endgroup$
    – Jan-Magnus Økland
    Dec 3 '18 at 18:21










  • $begingroup$
    Thanks for the detailed reply! However, we haven't encounterer Gröbner basis yet. Do you think there is still another way to get it without using Gröbner basis?
    $endgroup$
    – Mashed Potato
    Dec 3 '18 at 22:05










  • $begingroup$
    @MashedPotato Have you learned other tools from elimination theory? It is probably possible to do this with resultants instead of Gröbner bases. A more brute-force approach might also be possible. From a plot of the parametrization, we can see that the curve has degree $6$ (a generic line intersects it in $6$ points). So we write down a degree $6$ polynomial with $28$ undetermined coefficients. We can impose linear conditions conditions on these coefficients by plugging in points we obtain by substituting various values of $t$ into the parametrization, e.g., $t=0$ says $(1,0)$ lies on $C$.
    $endgroup$
    – André 3000
    Dec 3 '18 at 23:00










  • $begingroup$
    You can also plug the parametrization directly into the degree $6$ polynomial in $X$ and $Y$ and get linear relations that way. But this naïve method seems quite hard and tedious! The thing is, once you have a proposed answer, it's easy to check by substituting in the parametrization and seeing that the equation holds.
    $endgroup$
    – André 3000
    Dec 3 '18 at 23:00








  • 1




    $begingroup$
    I'm only taking an introductory course so I think what will really work for me are the brute force methods. Thanks for your help!!
    $endgroup$
    – Mashed Potato
    Dec 4 '18 at 2:59














  • 1




    $begingroup$
    Another way to get rid of fractions is to homogenize to I=ideal(X-(-t^6+15*t^4*u^2-15*t^2*u^4+u^6),Y-(-4*t^3*u+4*t*u^3)*(t^2+u^2),Z-(t^2+u^2)^3) and after taking the gröbner basis, set $Z=1$ in the resulting $16Y^6+4X^4Z^2-24Y^4Z^2-4X^2Z^4+9Y^2Z^4$. (+1)
    $endgroup$
    – Jan-Magnus Økland
    Dec 3 '18 at 18:21










  • $begingroup$
    Thanks for the detailed reply! However, we haven't encounterer Gröbner basis yet. Do you think there is still another way to get it without using Gröbner basis?
    $endgroup$
    – Mashed Potato
    Dec 3 '18 at 22:05










  • $begingroup$
    @MashedPotato Have you learned other tools from elimination theory? It is probably possible to do this with resultants instead of Gröbner bases. A more brute-force approach might also be possible. From a plot of the parametrization, we can see that the curve has degree $6$ (a generic line intersects it in $6$ points). So we write down a degree $6$ polynomial with $28$ undetermined coefficients. We can impose linear conditions conditions on these coefficients by plugging in points we obtain by substituting various values of $t$ into the parametrization, e.g., $t=0$ says $(1,0)$ lies on $C$.
    $endgroup$
    – André 3000
    Dec 3 '18 at 23:00










  • $begingroup$
    You can also plug the parametrization directly into the degree $6$ polynomial in $X$ and $Y$ and get linear relations that way. But this naïve method seems quite hard and tedious! The thing is, once you have a proposed answer, it's easy to check by substituting in the parametrization and seeing that the equation holds.
    $endgroup$
    – André 3000
    Dec 3 '18 at 23:00








  • 1




    $begingroup$
    I'm only taking an introductory course so I think what will really work for me are the brute force methods. Thanks for your help!!
    $endgroup$
    – Mashed Potato
    Dec 4 '18 at 2:59








1




1




$begingroup$
Another way to get rid of fractions is to homogenize to I=ideal(X-(-t^6+15*t^4*u^2-15*t^2*u^4+u^6),Y-(-4*t^3*u+4*t*u^3)*(t^2+u^2),Z-(t^2+u^2)^3) and after taking the gröbner basis, set $Z=1$ in the resulting $16Y^6+4X^4Z^2-24Y^4Z^2-4X^2Z^4+9Y^2Z^4$. (+1)
$endgroup$
– Jan-Magnus Økland
Dec 3 '18 at 18:21




$begingroup$
Another way to get rid of fractions is to homogenize to I=ideal(X-(-t^6+15*t^4*u^2-15*t^2*u^4+u^6),Y-(-4*t^3*u+4*t*u^3)*(t^2+u^2),Z-(t^2+u^2)^3) and after taking the gröbner basis, set $Z=1$ in the resulting $16Y^6+4X^4Z^2-24Y^4Z^2-4X^2Z^4+9Y^2Z^4$. (+1)
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– Jan-Magnus Økland
Dec 3 '18 at 18:21












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Thanks for the detailed reply! However, we haven't encounterer Gröbner basis yet. Do you think there is still another way to get it without using Gröbner basis?
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– Mashed Potato
Dec 3 '18 at 22:05




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Thanks for the detailed reply! However, we haven't encounterer Gröbner basis yet. Do you think there is still another way to get it without using Gröbner basis?
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– Mashed Potato
Dec 3 '18 at 22:05












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@MashedPotato Have you learned other tools from elimination theory? It is probably possible to do this with resultants instead of Gröbner bases. A more brute-force approach might also be possible. From a plot of the parametrization, we can see that the curve has degree $6$ (a generic line intersects it in $6$ points). So we write down a degree $6$ polynomial with $28$ undetermined coefficients. We can impose linear conditions conditions on these coefficients by plugging in points we obtain by substituting various values of $t$ into the parametrization, e.g., $t=0$ says $(1,0)$ lies on $C$.
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– André 3000
Dec 3 '18 at 23:00




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@MashedPotato Have you learned other tools from elimination theory? It is probably possible to do this with resultants instead of Gröbner bases. A more brute-force approach might also be possible. From a plot of the parametrization, we can see that the curve has degree $6$ (a generic line intersects it in $6$ points). So we write down a degree $6$ polynomial with $28$ undetermined coefficients. We can impose linear conditions conditions on these coefficients by plugging in points we obtain by substituting various values of $t$ into the parametrization, e.g., $t=0$ says $(1,0)$ lies on $C$.
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– André 3000
Dec 3 '18 at 23:00












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You can also plug the parametrization directly into the degree $6$ polynomial in $X$ and $Y$ and get linear relations that way. But this naïve method seems quite hard and tedious! The thing is, once you have a proposed answer, it's easy to check by substituting in the parametrization and seeing that the equation holds.
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– André 3000
Dec 3 '18 at 23:00






$begingroup$
You can also plug the parametrization directly into the degree $6$ polynomial in $X$ and $Y$ and get linear relations that way. But this naïve method seems quite hard and tedious! The thing is, once you have a proposed answer, it's easy to check by substituting in the parametrization and seeing that the equation holds.
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– André 3000
Dec 3 '18 at 23:00






1




1




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I'm only taking an introductory course so I think what will really work for me are the brute force methods. Thanks for your help!!
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– Mashed Potato
Dec 4 '18 at 2:59




$begingroup$
I'm only taking an introductory course so I think what will really work for me are the brute force methods. Thanks for your help!!
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– Mashed Potato
Dec 4 '18 at 2:59


















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