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Question Number 42340 by ajfour last updated on 23/Aug/18

Commented by ajfour last updated on 23/Aug/18

$F i n d r i n t e r m s o f R a n d a .$
Commented by MJS last updated on 24/Aug/18

$sorry I^{'} m on holidays and {don}^{'} t have internet$ $all the time . . .$
Commented by ajfour last updated on 25/Aug/18

$n o p r o b l e m s i r . e n j o y!$
	Answered by ajfour last updated on 25/Aug/18

	$e q . o f p a r a b o l a y = \frac{x^{2}}{4 a}$ $e q . o f s m a l l e r c i r c l e$ $x^{2} + {(y - 2 R + r)}^{2} = r^{2}$ $t h e s m a l l c i r c l e a n d p a r a b o l a$ $h a v e o n l y o n e c o m m o n v a l u e$ $o f x^{2}; h e n c e$ $x^{2} + {(\frac{x^{2}}{4 a} - 2 R + r)}^{2} = r^{2}$ $l e t x^{2} = t; t h e n$ $\frac{t^{2}}{16 a^{2}} + [1 - \frac{2 (2 R - r)}{4 a}] t + 2 R (2 R - 2 r) = 0$ $f o r j u s t o n e v a l u e o f t$ $\Rightarrow \frac{{[4 a - 2 (2 R - r)]}^{2}}{16 a^{2}} = \frac{8 R (2 R - 2 r)}{16 a^{2}}$ $\Rightarrow 4 a^{2} + {(2 R - r)}^{2} - 4 a (2 R - r)$ $= 4 R^{2} - 4 r R$ $\Rightarrow r^{2} + 4 a r + 4 a^{2} - 8 a R = 0$ $\Rightarrow r = - 2 a + \sqrt{4 a^{2} - 4 a^{2} + 8 a R}$ $\Rightarrow r = \sqrt{8 a R} - 2 a$ $f o r r = R$ $R^{2} + 4 a^{2} + 4 a R = 8 a R$ $\Rightarrow R = 2 a .$
	Commented by Tawa1 last updated on 25/Aug/18

	$Do you have your own textbook sir ? ? ? . Expecially all those your$ $diagrams . I will love to learn them sir . God bless you sir .$ $You, sir tammy and sir mrW 3 are genius . Keep the good work going sir .$ $God bless you sir .$
	Commented by ajfour last updated on 25/Aug/18

	$s e l f c r e a t e d q u e s t i o n s$ $p e r t a i n s t o m i x t u r e o f t o p i c s$ $i n c l u d i n g a p p l i c a t i o n o f d e r i a t i v e s,$ $c o o r d i n a t e g e o m e t r y, q u a d r a t i c,$ $g e o m e t r y, a n d l i k e . .$
	Commented by Tawa1 last updated on 25/Aug/18

	$Sir the drawings are what topic$
	Commented by Tawa1 last updated on 25/Aug/18

	$Wow! . That is great . . .$
	Commented by ajfour last updated on 25/Aug/18

	$y o u c a n s a v e m y q u e s t i o n s f r o m$ $t h i s t i n k u t a r a f o r u m (t o p i c - g e o m e t r y$ $o r c o o r d i n a t e g e o m e t r y) .$
	Commented by Tawa1 last updated on 25/Aug/18

	$My phone is format, i will start saving from today . And if you can share$ $me some past . i would appreciate it sir$
	Answered by MrW3 last updated on 25/Aug/18

	$C (0, R)$ $I (0, h)$ $B (u, v)$ $e q n . o f p a r a b o l a :$ $y = \frac{x^{2}}{4 a}$ $\Rightarrow \frac{d y}{d x} = \frac{x}{2 a}$ $e q n . o f s m a l l c i r c l e :$ $x^{2} + {(y - h)}^{2} = r^{2}$ $\Rightarrow 2 x + 2 (y - h) \frac{d y}{d x} = 0$ $\Rightarrow 2 u + 2 (v - h) \frac{u}{2 a} = 0$ $\Rightarrow v = h - 2 a$ $v = \frac{u^{2}}{4 a} \Rightarrow u^{2} = 4 a (h - 2 a)$ $h + r = 2 R \Rightarrow h = 2 R - r$ $u^{2} + {(v - h)}^{2} = r^{2}$ $\Rightarrow 4 a (h - 2 a) + {(h - 2 a - h)}^{2} = r^{2}$ $\Rightarrow 4 a h - 4 a^{2} = r^{2}$ $\Rightarrow 4 a (2 R - r) - 4 a^{2} = r^{2}$ $\Rightarrow 8 a R - 4 a r - 4 a^{2} = r^{2}$ $\Rightarrow r^{2} + 4 a r - 4 a (2 R - a) = 0$ $\Rightarrow r = \frac{- 4 a + \sqrt{16 a^{2} + 4 \times 4 a (2 R - a)}}{2}$ $\Rightarrow r = 2 (\sqrt{2 a R} - a)$ $s u c h t h a t r = R,$ $r = 2 (\sqrt{2 a R} - a) = R$ $2 \sqrt{2 a R} = R + 2 a$ ${(2 a - R)}^{2} = 0$ $\Rightarrow a = \frac{R}{2}$
	Commented by ajfour last updated on 25/Aug/18

	$t h a n k y o u S i r, t o o g o o d;$
	Commented by Tawa1 last updated on 24/Aug/18

	$God bless you sir .$ $Sir which topic can i learn diagrams like sir Ajfour always sent$
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