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Question Number 186616 by ajfour last updated on 07/Feb/23

	Answered by ajfour last updated on 07/Feb/23
	$say m=1 & forget the tangent. C(h,k)≡(1+p, r) let circle touches cubic curve at Q(q, q^3 −q−c) ⇒ (1+p−q)^2 +(q^3 −q−c−r)^2 =r^2 further, 3q^2 −1=((1+p−q)/(q^3 −q−c−r)) ⇒ p=q−1+(3q^2 −1)(q^3 −q−c−r) ⇒ {1+(3q^2 −1)^2 }(q^3 −q−c−r)^2 =r^2 lrt 1+(3q^2 −1)^2 =t^2 t(q^3 −q−c−r)=r p=q−1+r q=p+1−r 1+{3(p+1−r)^2 −1}^2 =t^2 ⇒(p+1−r)^2 =(1/3)±(1/3)(√(t^2 −1)) p^2 +2(1−r)p+(1−r)^2 =T p+c+2(1−r)p^2 +p(1−r)^2 =pT & 4(1−r)^2 p+2(1−r)p^2 +2(1−r)^3 =2(1−r)T subtracting {3(1−r)^2 −1+T}p =2(1−r)T+c−2(1−r)^3 ⇒ {2(1−r)T+c−2(1−r)^3 }^2 +2(1−r){2(1−r)T+c−2(1−r)^3 } ×{3(1−r)^2 −1+T} ={T−(1−r)^2 }{3(1−r)^2 −1+T}^2 say 1−r=R {2RT+c−2R^3 }^2 +2R{2RT+c−2R^3 }{T+3R^2 −1} =(T−R^2 )(T+3R^2 −1)^2 ⇒ T^( 3) −R^2 T^( 2) +2(3R^2 −1)T^( 2) −2R^2 (3R^2 −1)T+(3R^2 −1)^2 T −R^2 (3R^2 −1)^2 =4R^2 T^( 2) −4R(c−2R^3 )T+(c−2R^3 )^2 +4R^2 T^( 2) +4R^2 (3R^2 −1)T +2R(c−2R^3 )T+2R(c−2R^3 )(3R^2 −1) ⇒ if T=z, R=s z^3 −(3s^2 +2)z^2 +(1+2cs−13s^4 )z −{c−2s^3 +s(3s−1)}^2 =0 but 2s^3 −3s^2 +s−c=0 gives, for c=(1/3) s≈1.1989 now z^2 −(3s^2 +2)z+(1+2cs−13s^4 )=0 z=1+((3s^2 )/2)±(√((1+((3s^2 )/2))^2 +13s^4 −2cs−1)) p=r−1±(√T)=−s±(√z) .....$
	$s a y m = 1 & f o r g e t t h e t a n g e n t .$ $C (h, k) \equiv (1 + p, r)$ $l e t c i r c l e t o u c h e s c u b i c c u r v e a t$ $Q (q, q^{3} - q - c)$ $\Rightarrow {(1 + p - q)}^{2} + {(q^{3} - q - c - r)}^{2} = r^{2}$ $f u r t h e r,$ $3 q^{2} - 1 = \frac{1 + p - q}{q^{3} - q - c - r}$ $\Rightarrow p = q - 1 + (3 q^{2} - 1) (q^{3} - q - c - r)$ $\Rightarrow {1 + {(3 q^{2} - 1)}^{2}} {(q^{3} - q - c - r)}^{2} = r^{2}$ $l r t 1 + {(3 q^{2} - 1)}^{2} = t^{2}$ $t (q^{3} - q - c - r) = r$ $p = q - 1 + r$ $q = p + 1 - r$ $1 + {3 {(p + 1 - r)}^{2} - 1}^{2} = t^{2}$ $\Rightarrow {(p + 1 - r)}^{2} = \frac{1}{3} \pm \frac{1}{3} \sqrt{t^{2} - 1}$ $p^{2} + 2 (1 - r) p + {(1 - r)}^{2} = T$ $p + c + 2 (1 - r) p^{2} + p {(1 - r)}^{2} = p T$ $&$ $4 {(1 - r)}^{2} p + 2 (1 - r) p^{2} + 2 {(1 - r)}^{3}$ $= 2 (1 - r) T$ $s u b t r a c t i n g$ ${3 {(1 - r)}^{2} - 1 + T} p$ $= 2 (1 - r) T + c - 2 {(1 - r)}^{3}$ $\Rightarrow$ ${2 (1 - r) T + c - 2 {(1 - r)}^{3}}^{2}$ $+ 2 (1 - r) {2 (1 - r) T + c - 2 {(1 - r)}^{3}}$ $\times {3 {(1 - r)}^{2} - 1 + T}$ $= {T - {(1 - r)}^{2}} {3 {(1 - r)}^{2} - 1 + T}^{2}$ $s a y 1 - r = R$ ${2 R T + c - 2 R^{3}}^{2}$ $+ 2 R {2 R T + c - 2 R^{3}} {T + 3 R^{2} - 1}$ $= (T - R^{2}) {(T + 3 R^{2} - 1)}^{2}$ $\Rightarrow$ $T^{3} - R^{2} T^{2} + 2 (3 R^{2} - 1) T^{2}$ $- 2 R^{2} (3 R^{2} - 1) T + {(3 R^{2} - 1)}^{2} T$ $- R^{2} {(3 R^{2} - 1)}^{2}$ $= 4 R^{2} T^{2} - 4 R (c - 2 R^{3}) T + {(c - 2 R^{3})}^{2}$ $+ 4 R^{2} T^{2} + 4 R^{2} (3 R^{2} - 1) T$ $+ 2 R (c - 2 R^{3}) T + 2 R (c - 2 R^{3}) (3 R^{2} - 1)$ $\Rightarrow i f T = z, R = s$ $z^{3} - (3 s^{2} + 2) z^{2} + (1 + 2 c s - 13 s^{4}) z$ $- {c - 2 s^{3} + s (3 s - 1)}^{2} = 0$ $b u t 2 s^{3} - 3 s^{2} + s - c = 0$ $g i v e s, f o r c = \frac{1}{3} s \approx 1.1989$ $n o w$ $z^{2} - (3 s^{2} + 2) z + (1 + 2 c s - 13 s^{4}) = 0$ $z = 1 + \frac{3 s^{2}}{2} \pm \sqrt{{(1 + \frac{3 s^{2}}{2})}^{2} + 13 s^{4} - 2 c s - 1}$ $p = r - 1 \pm \sqrt{T} = - s \pm \sqrt{z}$ $. . . . .$
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