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Question Number 50568 by ajfour last updated on 17/Dec/18

Commented by ajfour last updated on 17/Dec/18

$F i n d e q u a t i o n o f t h e p a r a b o l a i n$ $r e d, i f t h e r e a r e N s u c h s u r r o u n d i n g$ $p a r a b o l a s . I n f i g u r e a b o v e N = 6 .$
	Answered by mr W last updated on 17/Dec/18

	$l e t y = {a x}^{2} + R$ $θ = \frac{2 π}{2 n} = \frac{π}{n}$ $y = \frac{x}{\tan θ}$ ${a x}^{2} - \frac{x}{\tan θ} + R = 0$ $D = \frac{1}{\tan^{2} θ} - 4 a R = 0$ $a = \frac{1}{4 R \tan^{2} θ}$ $\Rightarrow e q n . o f p a r a b o l a$ $y = \frac{x^{2}}{4 R \tan^{2} \frac{π}{n}} + R$
	Commented by ajfour last updated on 17/Dec/18

	$V e r y N i c e S i r!$
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