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Question Number 64009 by ajfour last updated on 12/Jul/19

Commented by ajfour last updated on 12/Jul/19

$F i n d R i n t e r m s o f a a n d b .$
	Answered by mr W last updated on 12/Jul/19

	Commented by mr W last updated on 12/Jul/19

	$e q n . o f p a r a b o l a :$ $y = - {A x}^{2} + b$ $0 = - {A a}^{2} + b$ $\Rightarrow A = \frac{b}{a^{2}}$ $\Rightarrow y = b (1 - \frac{x^{2}}{a^{2}})$ $y^{'} = - \frac{2 b x}{a^{2}}$ $P (x_{P}, y_{P})$ $y_{P} = b (1 - \frac{x_{P}^{2}}{a^{2}})$ $\frac{1}{\tan θ} = \frac{a^{2}}{2 {b x}_{P}}$ $\Rightarrow x_{P} = \frac{a^{2} \tan θ}{2 b}$ $y_{P} + R \cos θ = R$ $\Rightarrow y_{P} = R (1 - \cos θ) = b (1 - \frac{x_{P}^{2}}{a^{2}})$ $\Rightarrow \frac{R}{b} (1 - \cos θ) = (1 - \frac{a^{2} \tan^{2} θ}{4 b^{2}})$ $w i t h ρ = \frac{R}{b}, λ = \frac{a}{b}$ $\Rightarrow ρ = \frac{1}{1 - \cos θ} (1 - \frac{λ^{2} \tan^{2} θ}{4}) . . . (i)$ ${(2 R)}^{2} = {(R + b - R)}^{2} + {(x_{P} + R \sin θ)}^{2}$ $4 R^{2} = b^{2} + {(\frac{a^{2} \tan θ}{2 b} + R \sin θ)}^{2}$ $\Rightarrow 4 ρ^{2} = 1 + {(\frac{λ^{2} \tan θ}{2} + ρ \sin θ)}^{2} . . . (i i)$ $p u t (i) i n t o (i i) t o g e t θ$ $e x a m p l e :$ $a = 10, b = 5 \Rightarrow λ = a / b = 2$ $\Rightarrow θ = 39.6628 °$ $\Rightarrow ρ = 1.3579$
	Commented by mr W last updated on 12/Jul/19

	Commented by ajfour last updated on 12/Jul/19

	$c o o l, S i r . W o n d e r f u l .$
	Answered by ajfour last updated on 12/Jul/19
	$y=Ax^2 +b 0=Aa^2 +b ⇒ A=−b/a^2 let parabola touch the circle on right at (h,k). k=b(1−(h^2 /a^2 )) (dy/dx)=2Ax = 2(−(b/a^2 ))h slope of normal tan θ=(a^2 /(2bh)) let centre of right circle be C(p,q) p=h+Rcos θ , q=k+Rsin θ Now q=R and p^2 +b^2 =4R^2 ⇒ R=b(1−(h^2 /a^2 ))+Rsin θ (h+Rcos θ)^2 +b^2 =4R^2 tan θ=(a^2 /(2bh)) ⇒ h=((a^2 cos θ)/(2bsin θ)) ⇒ R=b(1−((a^2 cos^2 θ)/(4b^2 sin^2 θ)))+Rsin θ ⇒ R=(b/((1−sin θ)))(1−((a^2 cos^2 θ)/(4b^2 sin^2 θ))) (((a^2 cos θ)/(2bsin θ))+Rcos θ)^2 =4R^2 −b^2 [((a^2 cos θ)/(2bsin θ))+((b(1−((a^2 cos^2 θ)/(4b^2 sin^2 θ)))cos θ)/(1−sin θ))]^2 =[((2b(1−((a^2 cos^2 θ)/(4b^2 sin^2 θ))))/(1−sin θ))]^2 −b^2 let sin θ=t ⇒ (1−t^2 )[(a^2 /(2bt))+(b/(1−t))(1−((a^2 (1−t^2 ))/(4b^2 t^2 )))]^2 =[((2b(1−((a^2 (1−t^2 ))/(4b^2 t^2 ))))/(1−t))]^2 −b^2 ⇒ (1−t^2 )[2a^2 b(1−t)t+b{4b^2 t^2 −a^2 +a^2 t^2 }]^2 +16b^6 (1−t)^2 t^4 = 4b^2 [4b^2 t^2 −a^2 +a^2 t^2 ]^2 ⇒ (1−t^2 )[(4b^2 −a^2 )t^2 +2a^2 t−a^2 ]^2 +16b^4 (1−t)^2 t^4 =4[(a^2 +4b^2 )t^2 −a^2 ]^2 And for a=2, b=1 (1−t^2 )(2t−1)^2 =(1−t)^2 t^4 +4(2t^2 −1)^2 ⇒ t=0.7698144 R=(b/((1−sin θ)))(1−((a^2 cos^2 θ)/(4b^2 sin^2 θ))) R=((2t^2 −1)/(t^2 (1−t))) = 1.35787 ■$
	$y = {A x}^{2} + b$ $0 = {A a}^{2} + b \Rightarrow A = - b / a^{2}$ $l e t p a r a b o l a t o u c h t h e c i r c l e o n$ $r i g h t a t (h, k) .$ $k = b (1 - \frac{h^{2}}{a^{2}})$ $\frac{d y}{d x} = 2 A x = 2 (- \frac{b}{a^{2}}) h$ $s l o p e o f n o r m a l \tan θ = \frac{a^{2}}{2 b h}$ $l e t c e n t r e o f r i g h t c i r c l e b e$ $C (p, q)$ $p = h + R \cos θ, q = k + R \sin θ$ $N o w q = R a n d$ $p^{2} + b^{2} = 4 R^{2}$ $\Rightarrow R = b (1 - \frac{h^{2}}{a^{2}}) + R \sin θ$ ${(h + R \cos θ)}^{2} + b^{2} = 4 R^{2}$ $\tan θ = \frac{a^{2}}{2 b h} \Rightarrow h = \frac{a^{2} \cos θ}{2 b \sin θ}$ $\Rightarrow R = b (1 - \frac{a^{2} \cos^{2} θ}{4 b^{2} \sin^{2} θ}) + R \sin θ$ $\Rightarrow R = \frac{b}{(1 - \sin θ)} (1 - \frac{a^{2} \cos^{2} θ}{4 b^{2} \sin^{2} θ})$ ${(\frac{a^{2} \cos θ}{2 b \sin θ} + R \cos θ)}^{2} = 4 R^{2} - b^{2}$ ${[\frac{a^{2} \cos θ}{2 b \sin θ} + \frac{b (1 - \frac{a^{2} \cos^{2} θ}{4 b^{2} \sin^{2} θ}) \cos θ}{1 - \sin θ}]}^{2}$ $= {[\frac{2 b (1 - \frac{a^{2} \cos^{2} θ}{4 b^{2} \sin^{2} θ})}{1 - \sin θ}]}^{2} - b^{2}$ $l e t \sin θ = t$ $\Rightarrow (1 - t^{2}) {[\frac{a^{2}}{2 b t} + \frac{b}{1 - t} (1 - \frac{a^{2} (1 - t^{2})}{4 b^{2} t^{2}})]}^{2}$ $= {[\frac{2 b (1 - \frac{a^{2} (1 - t^{2})}{4 b^{2} t^{2}})}{1 - t}]}^{2} - b^{2}$ $\Rightarrow (1 - t^{2}) {[2 a^{2} b (1 - t) t + b {4 b^{2} t^{2} - a^{2} + a^{2} t^{2}}]}^{2}$ $+ 16 b^{6} {(1 - t)}^{2} t^{4}$ $= 4 b^{2} {[4 b^{2} t^{2} - a^{2} + a^{2} t^{2}]}^{2}$ $\Rightarrow$ $(1 - t^{2}) {[(4 b^{2} - a^{2}) t^{2} + 2 a^{2} t - a^{2}]}^{2}$ $+ 16 b^{4} {(1 - t)}^{2} t^{4} = 4 {[(a^{2} + 4 b^{2}) t^{2} - a^{2}]}^{2}$ $A n d f o r a = 2, b = 1$ $(1 - t^{2}) {(2 t - 1)}^{2} = {(1 - t)}^{2} t^{4} + 4 {(2 t^{2} - 1)}^{2}$ $\Rightarrow t = 0.7698144$ $R = \frac{b}{(1 - \sin θ)} (1 - \frac{a^{2} \cos^{2} θ}{4 b^{2} \sin^{2} θ})$ $R = \frac{2 t^{2} - 1}{t^{2} (1 - t)} = 1.35787 ◼$
	Commented by mr W last updated on 13/Jul/19

	$t h a n k s s i r!$ $i t s e e m s t h a t o n e c a n n o t g e t a f i n a l$ $e q u a t i o n o n l y f o r R .$
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