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Question Number 35646 by ajfour last updated on 21/May/18

Commented by ajfour last updated on 21/May/18

$F i n d t h e e q u a t i o n o f t h e p a r a b o l a$ $t h a t t o u c h e s t h e t h r e e c i r c l e s; (t h e$ $l a r g e r c i r c l e a t t h e v e r t e x) .$
Commented by tanmay.chaudhury50@gmail.com last updated on 22/May/18
$vertex=(0,−R) the centre of other two circle{(R+r)sinθ,− (R+r)cosθ} and{−(R+r)sinθ,−(R+r)cosθ} if we can find the points where the parabola touches the small circle then the equation of parabola can be found the eauation of parabola x^2 =−4A(y+R) this equation passes through vertex and two cintact point pls comment....$
$v e r t e x = (0, - R)$ $t h e c e n t r e o f o t h e r t w o c i r c l e {(R + r) s i n θ, -$ $(R + r) c o s θ} a n d {- (R + r) s i n θ, - (R + r) c o s θ}$ $i f w e c a n f i n d t h e p o i n t s w h e r e t h e p a r a b o l a$ $t o u c h e s t h e s m a l l c i r c l e t h e n t h e e q u a t i o n o f$ $p a r a b o l a c a n b e f o u n d$ $t h e e a u a t i o n o f p a r a b o l a$ $x^{2} = - 4 A (y + R)$ $t h i s e q u a t i o n p a s s e s t h r o u g h v e r t e x a n d t w o$ $c i n t a c t p o i n t$ $p l s c o m m e n t . . . .$
	Answered by ajfour last updated on 22/May/18

	$L e t r e q u i r e d e q . o f p a r a b o l a b e$ $y = {A x}^{2} - R$ $s l o p e o f c o m m o n t a n g e n t t o$ $s m a l l c i r c l e o n r i g h t a n d p a r a b o l a i s$ $m = - (\frac{x - (R + r) \sin θ}{y + (R + r) \cos θ}) = 2 A x . . (i)$ $a n d$ ${[y + (R + r) \cos θ]}^{2} = r^{2} - {[x - (R + r) \sin θ]}^{2}$ $. . . . . . . (i i)$ $u s i n g (i i) i n (i)$ $4 A^{2} x^{2} [r^{2} - {[x - (R + r) \sin θ]}^{2}$ $= {[x - (R + r) \sin θ]}^{2}$ $l e t x - (R + r) \sin θ = t$ $4 A^{2} {[t + (R + r) \sin θ]}^{2} (r^{2} - t^{2}) - t^{2} = 0$ $. . . . . . . . (I)$ $b e c a u s e o f t a n g e n c y, t w o$ $r o o t s a r e r e a l a n d e q u a l w h i l e$ $t h e o t h e r t w o a r e n o n r e a l . .$ $l e t r e a l r o o t s b e = a$ $w h i l e c o m p l e x o n e s p \pm i q; t h e n$ $- 4 A^{2} {(t - a)}^{2} (t - p + i q) (t - p - i q) = 0$ $\Rightarrow 4 A^{2} {(t - a)}^{2} (t^{2} - 2 p t + p^{2} - q^{2}) = 0$ $. . . . . . . (I I)$ $c o m p a r i n g c o e f f i c i e n t s o f$ $q u a d r a t i c t e r m t^{2} i n (I) a n d (I I),$ $4 A^{2} r^{2} - 4 A^{2} {(R + r)}^{2} \sin^{2} θ - 1$ $= 4 A^{2} [a^{2} + (p^{2} - q^{2}) + 4 a p]$ $b y c o m p a r i n g c o e f f . o f t^{3}$ $- 8 A^{2} (R + r) \sin θ = - 8 {p A}^{2} - 8 {a A}^{2}$ $\Rightarrow a + p = (R + r) \sin θ$ $b y c o m p a r i n g c o e f f . o f t$ $8 A^{2} r^{2} (R + r) \sin θ = 4 A^{2} [- 2 a^{2} p - 2 a (p^{2} - q^{2})]$ $. . . . . . . .$ $s h a l l h a v e t o c h a n g e t h e m e t h o d,$ $s o m e o n e h e l p p l e a s e!$
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