Show that x +y + z =0 is the only Asymptote of the curve x3 + y3 - 3axy =0
Answers
Answer:
The homogeneous equation is x3+y3−3axyz=0 . Setting z=0 , we get x3+y3=0 , so there are three improper points: (1:−1:0) , (1:ω:0) and 1:ω2:0) , where ω is a nonreal cube root of −1 .
However you seem only interested in the real asymptote. The partial derivatives of the homogeneous polynomial are
∂F∂x=3x2−3yz,∂F∂y=3y2−3axz,∂F∂z=−3axy
The values at the point (1:−1:0) are 3,3,3a . The asymptote is the tangent line at the point, therefore
3x+3y+3az=0
that can be written, in nonhomogeneous coordinates,
x+y+a=0
How can I approach drawing the curve for x^3+y^3-3axy=0 (folium of Descartes)?
What is the x^3+y^3-3axy maxima and minima solve?
What is the equation of the asymptote of x³+y³=3axy?
Let y=mx+c be the equation of asymptote.
x^3+m^3x^3+… =3amx^2
Ignore lower degree terms .m^3=-1 or m=-1
y+x=c
Go back to the equation.
(x+y)^3–3xy(x+y)=3axy
Put x+y=c
c^3–3cxy~3axy
c^3 is very small in comparison, ignore.
That gives c=-a
x+y+a=0 is the equation of asymptote.