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Question Number 155720 by ajfour last updated on 03/Oct/21

Commented by ajfour last updated on 03/Oct/21

$I f t h e p o l y n o m i a l c u r v e i s$ $y = x (x^{2} - p^{2}), f i n d r (p) .$
	Answered by TheSupreme last updated on 03/Oct/21

	$f (x) = x (x - p) (x + p)$ $P (x, x (x^{2} - p^{2}))$ $P (r (1 + c o s (a r c t a n (f^{'} (P)), r (1 - s i n (a r c t a n (f^{'} (P))$ $f^{'} (P) = (3 x^{2} - p^{2})$
	Answered by mr W last updated on 03/Oct/21
	$touching point A(a,b) with b=a(a^2 −p^2 ) tan θ=y′=3x^2 −p^2 =3a^2 −p^2 a=r(1+sin θ) b=r(1−cos θ) (b/a)=((1−cos θ)/(1+sin θ)) a^2 −p^2 =((1−cos θ)/(1+sin θ)) ⇒tan θ−((3(1−cos θ))/(1+sin θ))=2p^2 { ((p=(√((1/2)[tan θ−((3(1−cos θ))/(1+sin θ))])))),((r=(1/(1+sin θ))(√((1/2)[tan θ−((1−cos θ)/(1+sin θ))])))) :}$
	$t o u c h i n g p o i n t A (a, b) w i t h$ $b = a (a^{2} - p^{2})$ $\tan θ = y^{'} = 3 x^{2} - p^{2} = 3 a^{2} - p^{2}$ $a = r (1 + \sin θ)$ $b = r (1 - \cos θ)$ $\frac{b}{a} = \frac{1 - \cos θ}{1 + \sin θ}$ $a^{2} - p^{2} = \frac{1 - \cos θ}{1 + \sin θ}$ $\Rightarrow \tan θ - \frac{3 (1 - \cos θ)}{1 + \sin θ} = 2 p^{2}$ ${\begin{cases} p = \sqrt{\frac{1}{2} [\tan θ - \frac{3 (1 - \cos θ)}{1 + \sin θ}]} \\ r = \frac{1}{1 + \sin θ} \sqrt{\frac{1}{2} [\tan θ - \frac{1 - \cos θ}{1 + \sin θ}]} \end{cases}$
	Commented by mr W last updated on 03/Oct/21

	Commented by mr W last updated on 03/Oct/21

	Commented by mr W last updated on 03/Oct/21

	Answered by ajfour last updated on 01/Nov/21
	$x=r+rcos θ y=c=r−rsin θ ; p=1 ⇒ c=r(1−2sin φcos φ) 2φ=90°−θ ⇒ r(1−cos 2φ)=x(x^2 −p^2 ) r(1+cos θ)=x 2rsin^2 ((π/4)−(θ/2))=2rcos^2 (θ/2)(4r^2 cos^4 (θ/2)−p^2 ) say (θ/2)=δ (cos δ−sin δ)^2 =cos^2 δ(4r^2 cos^4 δ−p^2 ) ⇒ 2(1−sin θ) =(1+cos θ)[((4c^2 )/((1−sin θ)^2 ))−1] 2(1−sin θ)^3 =(1+cos θ){4c^2 −(1−sin θ)^2 } ......$
	$x = r + r \cos θ$ $y = c = r - r \sin θ; p = 1$ $\Rightarrow c = r (1 - 2 \sin ϕ \cos ϕ)$ $2 ϕ = 90 ° - θ$ $\Rightarrow r (1 - \cos 2 ϕ) = x (x^{2} - p^{2})$ $r (1 + \cos θ) = x$ $2 r \sin^{2} (\frac{π}{4} - \frac{θ}{2}) = 2 r \cos^{2} \frac{θ}{2} (4 r^{2} \cos^{4} \frac{θ}{2} - p^{2})$ $s a y \frac{θ}{2} = δ$ ${(\cos δ - \sin δ)}^{2} = \cos^{2} δ (4 r^{2} \cos^{4} δ - p^{2})$ $\Rightarrow$ $2 (1 - \sin θ)$ $= (1 + \cos θ) [\frac{4 c^{2}}{{(1 - \sin θ)}^{2}} - 1]$ $2 {(1 - \sin θ)}^{3}$ $= (1 + \cos θ) {4 c^{2} - {(1 - \sin θ)}^{2}}$ $. . . . . .$
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