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Question Number 200366 by ajfour last updated on 17/Nov/23

Commented by ajfour last updated on 17/Nov/23

$T h e r e d c i r c u l a r a r c l e n g t h i s e q u a l$ $t o b l u e a r c l e n g t h o f p a r a b o l a . F i n d$ $t h e e q u a t i o n o f p a r a b o l a i n t h e f o r m$ $y = {a x}^{2} + c .$
	Answered by Frix last updated on 17/Nov/23

	$Because of the intersections \Rightarrow c = \frac{\sqrt{3}}{2} - \frac{a}{4}$ $\underset{0}{\int^{\frac{1}{2}}} \sqrt{{(\frac{d y}{d x})}^{2} + 1} d x = \frac{π}{6}$ $\frac{\sqrt{a^{2} + 1}}{4} + \frac{\ln (a + \sqrt{a^{2} + 1})}{4 a} = \frac{π}{6}$ $a \approx . 543171783$ $c \approx . 730232458$
	Commented by ajfour last updated on 17/Nov/23

	$T h a n k s!$
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