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Question Number 187362 by ajfour last updated on 16/Feb/23

Commented by ajfour last updated on 16/Feb/23

$b l u e c u r v e : y = x^{3} - x$ $b l a c k o n e : y = x^{3} - x + k$ $F i n d t h e e q u a t i o n o f t h e s h o w n$ $c o m m o n t a n g e n t .$
	Answered by a.lgnaoui last updated on 17/Feb/23
	$Blue[curve y_1 =x^3 −x the equation of tangente is (dy/dx)=3x^2 −1 y has mean[and max to x: x_(1m) =(1/( (√3))) and x_(1max) =((−1)/( (√3))) point M_(12) (((−1)/( (√3))),(2/( 3(√3)))) N_(12) ((1/( (√3))),((−2)/(3(√3)))) the red line tengente also to blue and black[curve y_2 =x^3 −x+k same[(x min,x max) (((−1)/( (√3))),(2/( (√3)))+k) ,((1/( (√3))),k−(2/( (√3)))) y_2 (0)=k the equation y=ax+b of tengente must verifie: M_1 (((−1)/( (√3))),(2/(3(√3)))) and N((1/( (√3))),k−(2/( (√3)))) { (((2/( (√3))) =((−1)/( (√3)))a+b (1))),((k−(2/( (√3)))=(a/( (√3)))+b (2))) :} (1)+(2) ⇒k=2b b=(k/2). (2)−(1) ⇒k−(4/( (√3)))=((2a)/( (√3))) a=(k/2)(√3) −2 •Equation of tengente: y=((k/2)(√3) −2)x+(k/2) with k=y_2 (1) ; k>1$
	$B l u e [c u r v e y_{1} = x^{3} - x$ $t h e e q u a t i o n o f t a n g e n t e i s$ $\frac{d y}{d x} = 3 x^{2} - 1 y h a s m e a n [a n d m a x t o x :$ $x_{1 m} = \frac{1}{\sqrt{3}} a n d x_{1 m a x} = \frac{- 1}{\sqrt{3}}$ $p o i n t M_{12} (\frac{- 1}{\sqrt{3}}, \frac{2}{3 \sqrt{3}}) N_{12} (\frac{1}{\sqrt{3}}, \frac{- 2}{3 \sqrt{3}})$ $t h e r e d l i n e t e n g e n t e a l s o t o$ $b l u e a n d b l a c k [c u r v e$ $y_{2} = x^{3} - x + k s a m e [(x m i n, x m a x)$ $(\frac{- 1}{\sqrt{3}}, \frac{2}{\sqrt{3}} + k), (\frac{1}{\sqrt{3}}, k - \frac{2}{\sqrt{3}})$ $y_{2} (0) = k$ $t h e e q u a t i o n y = a x + b o f t e n g e n t e$ $m u s t v e r i f i e :$ $M_{1} (\frac{- 1}{\sqrt{3}}, \frac{2}{3 \sqrt{3}}) a n d N (\frac{1}{\sqrt{3}}, k - \frac{2}{\sqrt{3}})$ ${\begin{cases} \frac{2}{\sqrt{3}} = \frac{- 1}{\sqrt{3}} a + b (1) \\ k - \frac{2}{\sqrt{3}} = \frac{a}{\sqrt{3}} + b (2) \end{cases}$ $(1) + (2) \Rightarrow k = 2 b b = \frac{k}{2} .$ $(2) - (1) \Rightarrow k - \frac{4}{\sqrt{3}} = \frac{2 a}{\sqrt{3}}$ $a = \frac{k}{2} \sqrt{3} - 2$ $∙ E q u a t i o n o f t e n g e n t e :$ $y = (\frac{k}{2} \sqrt{3} - 2) x + \frac{k}{2}$ $w i t h k = y_{2} (1); k > 1$
	Commented by a.lgnaoui last updated on 17/Feb/23

	$k \in R n o t s p e c i f i e d!$ $t e n g e n t e [h e r e d e p e n d a n t$ $o f y_{1}, y_{2}$
	Answered by ajfour last updated on 17/Feb/23

	$T a n g e n t$ $y = m x + \frac{k}{2} (s y m m e t r y!)$ $I n t e r s e c t i o n ? w i t h c u r v e$ $y = x^{3} - x = m x + \frac{k}{2}$ $x^{3} - x = m x + \frac{k}{2}$ $x^{3} - (m + 1) x - \frac{k}{2} = 0$ $t a n g e n c y \Rightarrow d o u b l e r o o t \Rightarrow D = 0$ $\frac{k^{2}}{16} = \frac{{(m + 1)}^{3}}{27}$ $\Rightarrow m = \frac{3}{4} {(4 k^{2})}^{1 / 3} - 1$ $H e n c e e q . c o m m o n t a n g e n t$ $y = (\frac{3}{4} {(4 k^{2})}^{1 / 3} - 1) x + \frac{k}{2}$
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