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Question Number 90252 by ajfour last updated on 22/Apr/20

Commented by ajfour last updated on 22/Apr/20

$I f b o t h e l l i p s e s h a v e t h e s a m e$ $s h a p e, f i n d, b / a . A l s o f i n d$ $c i r c u m r a d i u s i n t e r m s o f a .$
	Answered by ajfour last updated on 22/Apr/20

	$l e t e q . o f r e d e l l i p s e b e$ $\frac{{(x - h)}^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$ $f o r x = 0, y^{2} = b^{2}$ $\Rightarrow h^{2} = b^{2} (1 - \frac{b^{2}}{a^{2}})$ $h + b = a$ $\Rightarrow {(a - b)}^{2} = b^{2} - \frac{b^{4}}{a^{2}}$ $l e t \frac{a}{b} = μ$ ${(μ - 1)}^{2} = 1 - \frac{1}{μ^{2}}$ $\Rightarrow μ^{4} - 2 μ^{3} + 1 = 0$ $\Rightarrow (μ - 1) (μ^{3} - μ^{2} - μ - 1) = 0$ $\Rightarrow i f μ \neq 1,$ $μ \approx 1.8393$
	Commented by mr W last updated on 22/Apr/20

	$a s o l u t i o n e x i s t s :$ $μ^{4} - 2 μ^{3} + 1 = 0$ $(μ - 1) (μ^{3} - μ^{2} - μ - 1) = 0$ $μ = 1 \Rightarrow n o s o l u t i o n$ $μ^{3} - μ^{2} - μ - 1 = 0$ $\Rightarrow μ = \frac{1}{3} (1 + \sqrt[3]{19 - 3 \sqrt{33}} + \sqrt[3]{19 + 3 \sqrt{33}}) = 1.8393$
	Commented by mr W last updated on 22/Apr/20

	Commented by ajfour last updated on 22/Apr/20

	$y e s s i r, μ \approx 1.8393, t h a n k s,$ $c a n w e t r y t h e s e c o n d p a r t t h e n$
	Answered by ajfour last updated on 22/Apr/20
	$let center of red ellipse be origin. Eq. of the ellipse (x^2 /b^2 )+(y^2 /a^2 )=1 (dy/dx)=−(a^2 /b^2 )((x/y)) let eq. of circle be (x−p)^2 +y^2 =r^2 for x=−(2a−b) , y=0 ⇒ 2a−b+p=r ....(i) let (s,t) be point of tangency of red ellipse and circle. (s−p)^2 +a^2 (1−(s^2 /b^2 ))=r^2 ((a^2 /b^2 )−1)s^2 +2ps+r^2 −p^2 −a^2 =0 D=0 ⇒ p^2 =((a^2 /b^2 )−1)(r^2 −p^2 −a^2 ) and as p=b+r−2a ((a^2 /b^2 )−1)r^2 −p^2 ((a^2 /b^2 ))=a^2 ((a^2 /b^2 )−1) (μ^2 −1)r^2 −μ^2 (b+r−2a)^2 =a^2 (μ^2 −1) r^2 −4μ^2 (a−(b/2))r+a^2 (μ^2 −1) +4μ^2 (a−(b/2))^2 =0 let r=λa λ^2 −4μ^2 (1−(1/(2μ)))λ+(μ^2 −1) +4μ^2 (1−(1/(2μ)))^2 =0 λ=2μ^2 (1−(1/(2μ)))−(√(4μ^2 (μ^2 −1)(1−(1/(2μ)))^2 −(μ^2 −1))) λ=μ(2μ−1)−(√((μ^2 −1){(2μ−1)^2 −1})) with μ=1.8393 λ≈ 1.09074$
	$l e t c e n t e r o f r e d e l l i p s e b e$ $o r i g i n . E q . o f t h e e l l i p s e$ $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1$ $\frac{d y}{d x} = - \frac{a^{2}}{b^{2}} (\frac{x}{y})$ $l e t e q . o f c i r c l e b e$ ${(x - p)}^{2} + y^{2} = r^{2}$ $f o r x = - (2 a - b), y = 0$ $\Rightarrow 2 a - b + p = r . . . . (i)$ $l e t (s, t) b e p o i n t o f t a n g e n c y$ $o f r e d e l l i p s e a n d c i r c l e .$ ${(s - p)}^{2} + a^{2} (1 - \frac{s^{2}}{b^{2}}) = r^{2}$ $(\frac{a^{2}}{b^{2}} - 1) s^{2} + 2 p s + r^{2} - p^{2} - a^{2} = 0$ $D = 0$ $\Rightarrow p^{2} = (\frac{a^{2}}{b^{2}} - 1) (r^{2} - p^{2} - a^{2})$ $a n d a s p = b + r - 2 a$ $(\frac{a^{2}}{b^{2}} - 1) r^{2} - p^{2} (\frac{a^{2}}{b^{2}}) = a^{2} (\frac{a^{2}}{b^{2}} - 1)$ $(μ^{2} - 1) r^{2} - μ^{2} {(b + r - 2 a)}^{2}$ $= a^{2} (μ^{2} - 1)$ $r^{2} - 4 μ^{2} (a - \frac{b}{2}) r + a^{2} (μ^{2} - 1)$ $+ 4 μ^{2} {(a - \frac{b}{2})}^{2} = 0$ $l e t r = λ a$ $λ^{2} - 4 μ^{2} (1 - \frac{1}{2 μ}) λ + (μ^{2} - 1)$ $+ 4 μ^{2} {(1 - \frac{1}{2 μ})}^{2} = 0$ $λ = 2 μ^{2} (1 - \frac{1}{2 μ}) - \sqrt{4 μ^{2} (μ^{2} - 1) {(1 - \frac{1}{2 μ})}^{2} - (μ^{2} - 1)}$ $λ = μ (2 μ - 1) - \sqrt{(μ^{2} - 1) {{(2 μ - 1)}^{2} - 1}}$ $w i t h μ = 1.8393$ $λ \approx 1.09074$
	Commented by mr W last updated on 22/Apr/20

	$v e r y n i c e s o l u t i o n s i r! e x a c t!$
	Commented by ajfour last updated on 22/Apr/20

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