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Question Number 157712 by ajfour last updated on 26/Oct/21

Commented by ajfour last updated on 26/Oct/21

$b l u e c u r v e i s y = x^{3} - x$ $F i n d e q . o f p a r a b o l a .$ $T h e l i n e s a r e$ $x = p w i t h p^{3} - p = c$ $y = x + c a n d y = c$
Commented by ajfour last updated on 26/Oct/21

$p a r a b o l a y = - {a x}^{2} + b x + k$ $t o u c h e s y = x$ $p a s s e s t h r o u g h (0, c) a n d$ $(p, c) . H e n c e u n i q u e!$
	Answered by ajfour last updated on 26/Oct/21
	$y=c+q^2 −A(x−q)^2 and p=2q so 8q^3 =2q+c x+c=c−Ax^2 +2Aqx+(1−A)q^2 Ax^2 +(2Aq−1)x+(1−A)q^2 =0 touches y=x+c hence (2Aq−1)^2 =4(1−A)q^2 ...(i) & 8q^3 =2q+c ...(ii) let q=s+t (2As+2At−1)^2 =4(1−A)(s+t)^2 if 2At+2As−1=m(s+t) ⇒ s+t=(1/(2A−m)) & m^2 =4(1−A) ⇒ A=1−(m^2 /4) (8/((2A−m)^3 ))=(2/((2A−m)))+c (8/((2−(m^2 /2)−m)^3 ))=(2/((2−(m^2 /2)−m)))+c 8=(2−(m^2 /2)−m)^2 +c(2−(m^2 /2)−m)^3 let m=2k 2=(1−k^2 −k)^2 +2c(1−k^2 −k)^3 2={(5/4)−(k−(1/2))^2 }^2 +2c{(5/4)−(k−(1/2))^2 }^3 let (k−(1/2))^2 =λ 2=((25)/(16))−((5λ)/2)+λ^2 +2c(((125)/(64))−λ^3 −((15λ^2 )/4)+((75λ)/(16))) 2cλ^3 +(((15c)/2)−1)λ^2 +((5/2)−((75c)/8))λ +((28−250c)/(64))=0$
	$y = c + q^{2} - A {(x - q)}^{2}$ $a n d p = 2 q$ $s o 8 q^{3} = 2 q + c$ $x + c = c - {A x}^{2} + 2 A q x + (1 - A) q^{2}$ ${A x}^{2} + (2 A q - 1) x + (1 - A) q^{2} = 0$ $t o u c h e s y = x + c h e n c e$ ${(2 A q - 1)}^{2} = 4 (1 - A) q^{2} . . . (i)$ $&$ $8 q^{3} = 2 q + c . . . (i i)$ $l e t q = s + t$ ${(2 A s + 2 A t - 1)}^{2} = 4 (1 - A) {(s + t)}^{2}$ $i f 2 A t + 2 A s - 1 = m (s + t)$ $\Rightarrow s + t = \frac{1}{2 A - m}$ $& m^{2} = 4 (1 - A)$ $\Rightarrow A = 1 - \frac{m^{2}}{4}$ $\frac{8}{{(2 A - m)}^{3}} = \frac{2}{(2 A - m)} + c$ $\frac{8}{{(2 - \frac{m^{2}}{2} - m)}^{3}} = \frac{2}{(2 - \frac{m^{2}}{2} - m)} + c$ $8 = {(2 - \frac{m^{2}}{2} - m)}^{2} + c {(2 - \frac{m^{2}}{2} - m)}^{3}$ $l e t m = 2 k$ $2 = {(1 - k^{2} - k)}^{2} + 2 c {(1 - k^{2} - k)}^{3}$ $2 = {\frac{5}{4} - {(k - \frac{1}{2})}^{2}}^{2} + 2 c {\frac{5}{4} - {(k - \frac{1}{2})}^{2}}^{3}$ $l e t {(k - \frac{1}{2})}^{2} = λ$ $2 = \frac{25}{16} - \frac{5 λ}{2} + λ^{2} + 2 c (\frac{125}{64} - λ^{3} - \frac{15 λ^{2}}{4} + \frac{75 λ}{16})$ $2 c λ^{3} + (\frac{15 c}{2} - 1) λ^{2} + (\frac{5}{2} - \frac{75 c}{8}) λ$ $+ \frac{28 - 250 c}{64} = 0$
	Answered by mr W last updated on 27/Oct/21

	Commented by mr W last updated on 27/Oct/21

	$A (0, c)$ $B (p, c)$ $p^{3} - p - c = 0$ $i f c ⩾ \frac{2}{3 \sqrt{3}} :$ $p = \sqrt[3]{\sqrt{\frac{c^{2}}{4} - \frac{1}{27}} + \frac{c}{2}} - \sqrt[3]{\sqrt{\frac{c^{2}}{4} - \frac{1}{27}} - \frac{c}{2}}$ $i f c < \frac{2}{3 \sqrt{3}} :$ $p = \frac{2}{\sqrt{3}} \sin (- \frac{1}{3} \sin^{- 1} \frac{3 \sqrt{3} c}{2} + \frac{2 π}{3})$ $s a y p a r a b o l a i s$ $y = A {(x - \frac{p}{2})}^{2} + k$ $y^{'} = 2 A (x - \frac{p}{2})$ $1 = 2 A (0 - \frac{p}{2}) \Rightarrow A = - \frac{1}{p}$ $c = - \frac{1}{p} {(0 - \frac{p}{2})}^{2} + k \Rightarrow k = c + \frac{p}{4}$ $\Rightarrow y = - \frac{1}{p} {(x - \frac{p}{2})}^{2} + \frac{p}{4} + c$ $\Rightarrow y = - \frac{x (x - p)}{p} + c$
	Commented by mr W last updated on 27/Oct/21

	Commented by mr W last updated on 27/Oct/21

	Commented by ajfour last updated on 27/Oct/21

	$t h a n k y o u s i r!$
	Commented by Ahmed1hamouda last updated on 29/Oct/21

	$What applications do you use in thee$ $enginering solution$
	Commented by mr W last updated on 29/Oct/21

	$G r a p h e r$
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