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Question Number 186329 by ajfour last updated on 03/Feb/23

Answered by ajfour last updated on 03/Feb/23

Cubic curve: y=x^3 −x Parabola: x=h+(y−k)^2 x=h+(p−k)^2 p^3 −p=c ⇒ p^2 =2kp+x−h−k^2 say x−h−k^2 =t p^2 =2kp+t p+c=2kp^2 +tp 2kp^2 =(1−t)p+c=4k^2 p+2kt ⇒ p=((c−2kt)/(4k^2 +t−1)) =((c−mt)/(m^2 +t−1)) now, since p^2 =2kp+t (c−mt)^2 =m(c−mt)(m^2 +t−1) +t(t+m^2 −1)^2 ⇒ c^2 +m^2 t^2 −2cmt =cm^3 +cmt−cm−m^4 t−m^2 t^2 +m^2 t +t^3 +2(m^2 −1)t^2 +(m^2 −1)^2 t ⇒ t^3 −2t^2 +(3cm−m^2 +1)t +c(m^3 −m−c)=0 ....

Cubiccurve:y=x3−xParabola:x=h+(y−k)2x=h+(p−k)2p3−p=c⇒p2=2kp+x−h−k2sayx−h−k2=tp2=2kp+tp+c=2kp2+tp2kp2=(1−t)p+c=4k2p+2kt⇒p=c−2kt4k2+t−1=c−mtm2+t−1now,sincep2=2kp+t(c−mt)2=m(c−mt)(m2+t−1)+t(t+m2−1)2⇒c2+m2t2−2cmt=cm3+cmt−cm−m4t−m2t2+m2t+t3+2(m2−1)t2+(m2−1)2t⇒t3−2t2+(3cm−m2+1)t+c(m3−m−c)=0....

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