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Question Number 44763 by ajfour last updated on 04/Oct/18

Commented by ajfour last updated on 04/Oct/18

$E q u a t i o n o f e l l i p s e : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 .$ $F i n d r i n t e r m s o f a, b .$
	Answered by ajfour last updated on 04/Oct/18
	$let point of contact of circle and ellipse be (x_0 , y_0 ) let y_0 = bsin 𝛗 = r(1+sin 𝛉) ...(i) x_0 = acos 𝛗 = r(1+cos 𝛉) ...(ii) slope of common tangent: ((bcos 𝛗)/(−asin 𝛗)) = −cot 𝛉 ⇒ bcos 𝛗sin 𝛉 = asin 𝛗cos 𝛉 ..(iii) ⇒ tan 𝛗 = ((bsin θ)/(acos θ)) = (a/b)(((1+sin 𝛉)/(1+cos 𝛉))) we must have a double root for this eq. in θ ; let tan 𝛉 = m let tan (θ/2) = t , then 2t = m(1−t^2 ) and also get (b/a)×((2t)/(1−t^2 )) = (a/b)×((1+t^2 +2t)/2) ⇒ 4b^2 t = a^2 (1−t^2 )(1+t)^2 ⇒ 4b^2 t = a^2 (1+2t+t^2 −t^2 −2t^3 −t^4 ) let ((4b^2 )/a^2 )−2= 𝛒 (ρ > 0 if b(√2) > a ) ⇒ t^4 +2t^3 +ρt−1=0 this eq. must have two real roots corresponding to tan θ = m and two complex roots; say (mt^2 +2t−m)(t^2 +pt+q)=0 t^4 +2t^3 +ρt−1 ⇔ (t^2 +((2t)/m)−1)(t^2 +pt+q) ⇒ for t=0 −q = −1 ....(1) for t=1 2+𝛒 = (2/m)(1+p+q) ...(2) for t=−1 −2−𝛒 =−(2/m)(1−p+q) ...(3) from eq.(2) & eq.(3) we infer p = 0 and therefore from (2) (2/m)(1+0+1) = 2+𝛒 ⇒ m= tan θ = (4/(2+ρ)) = (a^2 /b^2 ) And as mt^2 +2t−m = 0 with appropriate t=tan (θ/2) being +ve and < 1 , we conclude t = −(1/m)+(√(1+(1/m^2 ))) and with m= (a^2 /b^2 ) t = −(b^2 /a^2 )+((√(a^4 +b^4 ))/a^2 ) ⇒ from (i) & (ii) (in the beginning), r^2 [(((1+sin θ)^2 )/b^2 )+(((1+cos θ)^2 )/a^2 )]=1 ⇒ r^2 = ((a^2 b^2 )/(a^2 (1+sin 𝛉)^2 +b^2 (1+cos 𝛉)^2 )) = ((a^2 b^2 )/(a^2 (1+((2t)/(1+t^2 )))^2 +b^2 (1+((1−t^2 )/(1+t^2 )))^2 )) { ((r^2 = ((a^2 b^2 (1+t^2 )^2 )/(a^2 (1+t)^4 +4b^2 )))),((with t=−(b^2 /a^2 )+((√(a^4 +b^4 ))/a^2 ))) :} As a check , if a=b=R we have t= (√2)−1 and r = ((ab(1+t^2 ))/(√(a^2 (1+t)^4 +4b^2 ))) ⇒ r = (((4−2(√2))R)/(2(√2))) = (((2−(√2))R)/(√2)) = ((2R)/((√2)(2+(√2)))) = (R/(1+(√2))) (as it should be !) .$
	$l e t p o i n t o f c o n t a c t o f c i r c l e$ $a n d e l l i p s e b e (x_{0}, y_{0})$ $l e t$ $y_{0} = b \sin ϕ = r (1 + \sin θ) . . . (i)$ $x_{0} = a \cos ϕ = r (1 + \cos θ) . . . (i i)$ $s l o p e o f c o m m o n t a n g e n t :$ $\frac{b \cos ϕ}{- a \sin ϕ} = - \cot θ$ $\Rightarrow b \cos ϕ \sin θ = a \sin ϕ \cos θ . . (i i i)$ $\Rightarrow \tan ϕ = \frac{b \sin θ}{a \cos θ} = \frac{a}{b} (\frac{1 + \sin θ}{1 + \cos θ})$ $w e m u s t h a v e a d o u b l e r o o t f o r$ $t h i s e q . i n θ; l e t \tan θ = m$ $l e t \tan \frac{θ}{2} = t, t h e n$ $2 t = m (1 - t^{2}) a n d a l s o g e t$ $\frac{b}{a} \times \frac{2 t}{1 - t^{2}} = \frac{a}{b} \times \frac{1 + t^{2} + 2 t}{2}$ $\Rightarrow 4 b^{2} t = a^{2} (1 - t^{2}) {(1 + t)}^{2}$ $\Rightarrow 4 b^{2} t = a^{2} (1 + 2 t + t^{2} - t^{2} - 2 t^{3} - t^{4})$ $l e t \frac{4 b^{2}}{a^{2}} - 2 = ρ (ρ > 0 i f b \sqrt{2} > a)$ $\Rightarrow t^{4} + 2 t^{3} + ρ t - 1 = 0$ $t h i s e q . m u s t h a v e t w o r e a l$ $r o o t s c o r r e s p o n d i n g t o \tan θ = m$ $a n d t w o c o m p l e x r o o t s;$ $s a y ({m t}^{2} + 2 t - m) (t^{2} + p t + q) = 0$ $t^{4} + 2 t^{3} + ρ t - 1 \Leftrightarrow (t^{2} + \frac{2 t}{m} - 1) (t^{2} + p t + q)$ $\Rightarrow f o r t = 0$ $- q = - 1 . . . . (1)$ $f o r t = 1$ $2 + ρ = \frac{2}{m} (1 + p + q) . . . (2)$ $f o r t = - 1$ $- 2 - ρ = - \frac{2}{m} (1 - p + q) . . . (3)$ $f r o m e q . (2) & e q . (3) w e i n f e r$ $p = 0 a n d t h e r e f o r e f r o m (2)$ $\frac{2}{m} (1 + 0 + 1) = 2 + ρ$ $\Rightarrow m = \tan θ = \frac{4}{2 + ρ} = \frac{a^{2}}{b^{2}}$ $A n d a s {m t}^{2} + 2 t - m = 0$ $w i t h a p p r o p r i a t e t = \tan \frac{θ}{2}$ $b e i n g + v e a n d < 1, w e c o n c l u d e$ $t = - \frac{1}{m} + \sqrt{1 + \frac{1}{m^{2}}}$ $a n d w i t h m = \frac{a^{2}}{b^{2}}$ $t = - \frac{b^{2}}{a^{2}} + \frac{\sqrt{a^{4} + b^{4}}}{a^{2}}$ $\Rightarrow$ $f r o m (i) & (i i) (i n t h e b e g i n n i n g),$ $r^{2} [\frac{{(1 + \sin θ)}^{2}}{b^{2}} + \frac{{(1 + \cos θ)}^{2}}{a^{2}}] = 1$ $\Rightarrow r^{2} = \frac{a^{2} b^{2}}{a^{2} {(1 + \sin θ)}^{2} + b^{2} {(1 + \cos θ)}^{2}}$ $= \frac{a^{2} b^{2}}{a^{2} {(1 + \frac{2 t}{1 + t^{2}})}^{2} + b^{2} {(1 + \frac{1 - t^{2}}{1 + t^{2}})}^{2}}$ ${\begin{cases} r^{2} = \frac{a^{2} b^{2} {(1 + t^{2})}^{2}}{a^{2} {(1 + t)}^{4} + 4 b^{2}} \\ w i t h t = - \frac{b^{2}}{a^{2}} + \frac{\sqrt{a^{4} + b^{4}}}{a^{2}} \end{cases}$ $A s a c h e c k, i f a = b = R$ $w e h a v e t = \sqrt{2} - 1$ $a n d r = \frac{a b (1 + t^{2})}{\sqrt{a^{2} {(1 + t)}^{4} + 4 b^{2}}}$ $\Rightarrow r = \frac{(4 - 2 \sqrt{2}) R}{2 \sqrt{2}} = \frac{(2 - \sqrt{2}) R}{\sqrt{2}}$ $= \frac{2 R}{\sqrt{2} (2 + \sqrt{2})} = \frac{R}{1 + \sqrt{2}}$ $(a s i t s h o u l d b e!) .$
	Commented by MrW3 last updated on 04/Oct/18

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