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Question Number 187059 by ajfour last updated on 13/Feb/23

Commented by ajfour last updated on 13/Feb/23

$B i g g e r c u r v e y = x^{3} - x - c$ $t h e o t h e r y = m {(x - \frac{1}{\sqrt{3}})}^{2} (x - s)$ $s a y t h e i r i n t e r s e c t i o n s b e a t$ $x = p, q, t$ $A l s o p^{3} = p + c$ $w e m u s t f i n d p i n s u r d s . . . e v e n i f$ $0 < c < \frac{2}{3 \sqrt{3}} ★$
	Answered by ajfour last updated on 13/Feb/23
	$y=x^3 −x Y=m(x−(1/( (√3))))^2 (x−s) =m(x^2 −((2x)/( (√3)))+(1/3))(x−s) y=Y=c=p^3 −p =m{p^3 −(s+(2/( (√3))))p^2 +(((2s)/( (√3)))+(1/3))p−(s/3)} ⇒(m−1)p^3 −m(s+(2/( (√3))))p^2 +{m(((2s)/( (√3)))+(1/3))+1}p−((ms)/3)=0 let m^2 (s+(2/( (√3))))^2 =3m(m−1)(((2s)/( (√3)))+(1/3)) +3(m−1) let s=−(1/(2(√3))) ⇒ m^2 =4(m−1) ⇒ (m−2)^2 =0 ⇒ m=2 Now c=2{p^3 −(((√3)p^2 )/2)+(1/(6(√3)))} and for c=(1/3) p^3 −(((√3)p^2 )/2)−(1/6)(1−(1/( (√3))))=0 let me check this...$
	$y = x^{3} - x$ $Y = m {(x - \frac{1}{\sqrt{3}})}^{2} (x - s)$ $= m (x^{2} - \frac{2 x}{\sqrt{3}} + \frac{1}{3}) (x - s)$ $y = Y = c = p^{3} - p$ $= m {p^{3} - (s + \frac{2}{\sqrt{3}}) p^{2} + (\frac{2 s}{\sqrt{3}} + \frac{1}{3}) p - \frac{s}{3}}$ $\Rightarrow (m - 1) p^{3} - m (s + \frac{2}{\sqrt{3}}) p^{2}$ $+ {m (\frac{2 s}{\sqrt{3}} + \frac{1}{3}) + 1} p - \frac{m s}{3} = 0$ $l e t$ $m^{2} {(s + \frac{2}{\sqrt{3}})}^{2} = 3 m (m - 1) (\frac{2 s}{\sqrt{3}} + \frac{1}{3})$ $+ 3 (m - 1)$ $l e t s = - \frac{1}{2 \sqrt{3}}$ $\Rightarrow m^{2} = 4 (m - 1)$ $\Rightarrow {(m - 2)}^{2} = 0 \Rightarrow m = 2$ $N o w$ $c = 2 {p^{3} - \frac{\sqrt{3} p^{2}}{2} + \frac{1}{6 \sqrt{3}}}$ $a n d f o r c = \frac{1}{3}$ $p^{3} - \frac{\sqrt{3} p^{2}}{2} - \frac{1}{6} (1 - \frac{1}{\sqrt{3}}) = 0$ $l e t m e c h e c k t h i s . . .$
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