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Question Number 142499 by ajfour last updated on 01/Jun/21

Commented by ajfour last updated on 01/Jun/21

$T h e r e d c i r c l e h a s r a d i u s c .$ $T h e b l u e p a i r o f l i n e s a r e$ $m u t u a l l y p e r p e n d i c u l a r .$ $L o c a t e A . (c < \frac{2}{3 \sqrt{3}})$
	Answered by mr W last updated on 01/Jun/21

	$A (p, p^{2})$ $B (- b, p^{2})$ $\frac{b}{p^{2}} = \frac{p^{2}}{p}$ $\Rightarrow b = p^{3}$ $b^{2} + {(p^{2})}^{2} = c^{2}$ ${(p^{2})}^{3} + {(p^{2})}^{2} = c^{2}$ ${(\frac{1}{p^{2}})}^{3} - \frac{1}{c^{2}} (\frac{1}{p^{2}}) - \frac{1}{c^{2}} = 0$ $\Rightarrow \frac{1}{p^{2}} = \sqrt[3]{\frac{1}{2 c^{2}} (1 + \sqrt{1 - \frac{4}{27 c^{2}}})} + \sqrt[3]{\frac{1}{2 c^{2}} (1 - \sqrt{1 - \frac{4}{27 c^{2}}})}$ $\Rightarrow p = \frac{1}{\sqrt{\sqrt[3]{\frac{1}{2 c^{2}} (1 + \sqrt{1 - \frac{4}{27 c^{2}}})} + \sqrt[3]{\frac{1}{2 c^{2}} (1 - \sqrt{1 - \frac{4}{27 c^{2}}})}}}$
	Commented by mr W last updated on 01/Jun/21

	Commented by mr W last updated on 01/Jun/21

	Commented by ajfour last updated on 01/Jun/21

	$T h a n k s S i r, b u t t h i s a g a i n$ ${d o e s n}^{'} t w o r k f o r 27 c^{2} < 4 .$
	Commented by mr W last updated on 01/Jun/21

	$y e s .$
	Answered by ajfour last updated on 01/Jun/21
	$let A(p,p^2 ) B(−(√(c^2 −p^4 )), p^2 ) (x−((p−(√(c^2 −p^4 )))/2))^2 +(y−p^2 )^2 =(1/4)(((p+(√(c^2 −p^4 )))/2))^2 this circle with AB as the horizontal diameter passes through origin; hence 4p^4 +4(((p−(√(c^2 −p^4 )))/2))^2 =(((p+(√(c^2 −p^4 )))/2))^2 ⇒ 4p^4 +p^2 +c^2 −p^4 −2p(√(c^2 −p^4 )) =(p^2 /4)+(c^2 /4)−(p^4 /4)+((p(√(c^2 −p^4 )))/2) ⇒13p^4 +3p^2 +3c^2 =9p(√(c^2 −p^4 )) let 9p(√(c^2 −p^4 ))=ap^2 +b(c^2 −p^4 ) ⇒ ((ap)/( (√(c^2 −p^4 ))))+((b(√(c^2 −p^4 )))/p)=9 at+(b/t)=9 at^2 −9t+b=0 t=(9/(2a))±(√(((81)/(4a^2 ))−b)) , & (13+b)p^4 +(3−a)p^2 +(3−b)c^2 =0 let b=−13 ⇒ (3−a)p^2 +16c^2 =0 at^2 =9t+13=((81)/(2a))+(√((((81)/(2a)))^2 +13×81)) ⇒ ((a(((16c^2 )/(a−3))))/(c^2 −(((16c^2 )/(a−3)))^2 ))=((81)/(2a))+(√((((81)/(2a)))^2 +13×81)) {a(((16c^2 )/(a−3)))−((81c^2 )/(2a))+((81)/(2a))(((16c^2 )/(a−3)))^2 }^2 ={c^2 −(((16c^2 )/(a−3)))^2 }^2 {(((81)/(2a)))^2 +13×81} ⇒ {2a^2 (a−3)(16c^2 )−81c^2 (a−3)^2 +81(16c^2 )^2 }^2 ={c^2 (a−3)^2 −(16c^2 )^2 }^2 {(81)^2 +13×81(2a)^2 } ....$
	$l e t A (p, p^{2})$ $B (- \sqrt{c^{2} - p^{4}}, p^{2})$ ${(x - \frac{p - \sqrt{c^{2} - p^{4}}}{2})}^{2} + {(y - p^{2})}^{2}$ $= \frac{1}{4} {(\frac{p + \sqrt{c^{2} - p^{4}}}{2})}^{2}$ $t h i s c i r c l e w i t h A B a s t h e$ $h o r i z o n t a l d i a m e t e r p a s s e s$ $t h r o u g h o r i g i n; h e n c e$ $4 p^{4} + 4 {(\frac{p - \sqrt{c^{2} - p^{4}}}{2})}^{2} = {(\frac{p + \sqrt{c^{2} - p^{4}}}{2})}^{2}$ $\Rightarrow 4 p^{4} + p^{2} + c^{2} - p^{4} - 2 p \sqrt{c^{2} - p^{4}}$ $= \frac{p^{2}}{4} + \frac{c^{2}}{4} - \frac{p^{4}}{4} + \frac{p \sqrt{c^{2} - p^{4}}}{2}$ $\Rightarrow 13 p^{4} + 3 p^{2} + 3 c^{2} = 9 p \sqrt{c^{2} - p^{4}}$ $l e t 9 p \sqrt{c^{2} - p^{4}} = {a p}^{2} + b (c^{2} - p^{4})$ $\Rightarrow \frac{a p}{\sqrt{c^{2} - p^{4}}} + \frac{b \sqrt{c^{2} - p^{4}}}{p} = 9$ $a t + \frac{b}{t} = 9$ ${a t}^{2} - 9 t + b = 0$ $t = \frac{9}{2 a} \pm \sqrt{\frac{81}{4 a^{2}} - b}, &$ $(13 + b) p^{4} + (3 - a) p^{2} + (3 - b) c^{2} = 0$ $l e t b = - 13 \Rightarrow$ $(3 - a) p^{2} + 16 c^{2} = 0$ ${a t}^{2} = 9 t + 13 = \frac{81}{2 a} + \sqrt{{(\frac{81}{2 a})}^{2} + 13 \times 81}$ $\Rightarrow \frac{a (\frac{16 c^{2}}{a - 3})}{c^{2} - {(\frac{16 c^{2}}{a - 3})}^{2}} = \frac{81}{2 a} + \sqrt{{(\frac{81}{2 a})}^{2} + 13 \times 81}$ ${a (\frac{16 c^{2}}{a - 3}) - \frac{81 c^{2}}{2 a} + \frac{81}{2 a} {(\frac{16 c^{2}}{a - 3})}^{2}}^{2}$ $= {c^{2} - {(\frac{16 c^{2}}{a - 3})}^{2}}^{2} {{(\frac{81}{2 a})}^{2} + 13 \times 81}$ $\Rightarrow$ ${2 a^{2} (a - 3) (16 c^{2}) - 81 c^{2} {(a - 3)}^{2}$ ${+ 81 {(16 c^{2})}^{2}}}^{2}$ $= {c^{2} {(a - 3)}^{2} - {(16 c^{2})}^{2}}^{2} {{(81)}^{2} + 13 \times 81 {(2 a)}^{2}}$ $. . . .$
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