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Question Number 176147 by peter frank last updated on 13/Sep/22

	Answered by mr W last updated on 13/Sep/22

	Commented by mr W last updated on 13/Sep/22

	$t h e t r a c k o f t h e b a l l i s a p a r a b o l a$ $t h r o u g h p o i n t s O, A, B, C . {i t}^{'} s e q n . i s$ $y = k x (x - R)$ $w e h a v e$ $a = k b (b - R) . . . (i)$ $b = k a (a - R) . . . (i i)$ $(i) / (i i) :$ $\frac{a}{b} = \frac{b (b - R)}{a (a - R)}$ $a^{3} - a^{2} R = b^{3} - b^{2} R$ $(b^{2} - a^{2}) R = b^{3} - a^{3} = (b - a) (a^{2} + b^{2} + a b)$ $\Rightarrow R = \frac{a^{2} + a b + b^{2}}{a + b} ✓$
	Commented by peter frank last updated on 14/Sep/22

	$thank you$
	Commented by Tawa11 last updated on 15/Sep/22

	$Great sir$
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