Question Number 71739 by Tony Lin last updated on 19/Oct/19

find the asymptote of folium of Descartes x^3 +y^3 =3axy, and a is a constant >0

Answered by Tony Lin last updated on 19/Oct/19

let y=tx ⇒x^3 +(tx)^3 =3atx^2 { ((x=((3at)/(1+t^3 )))),((y=((3at^2 )/(1+t^3 )))) :} x→±∞⇒t→−1 let the asymptote L: mx+b lim_(t→−1) (y/x)=−1=m lim_(t→−1) [y−(−x)] =lim_(x→−1) ((3at(t+1))/((t+1)(t^2 −t+1))) =−a =b therefore asymptote L:−x−a

Answered by MJS last updated on 19/Oct/19

x^3 +y^3 −3axy=0 turning x=((√2)/2)(x^∗ +y^∗ )∧y=((√2)/2)(x^∗ −y^∗ ) ((3(a+(√2)x^∗ ))/2)(y^∗ )^2 −(((3a−(√2)x^∗ )(x^∗ )^2 )/2)=0 ⇒ y^∗ =±((x^∗ (√(9a−3(√2)x^∗ )))/(3(√(a+(√2)x^∗ )))) denominator not defined for x^∗ =−(((√2)a)/2)+0y^∗ turning back p=((√2)/2)(x+y)∧q=((√2)/2)(x−y) ((√2)/2)(x+y)=−(((√2)a)/2) y=−x−a is the asymptote

Commented byMJS last updated on 19/Oct/19

there might be an easier path...