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Question Number 89192 by ajfour last updated on 16/Apr/20

Commented by Prithwish Sen 1 last updated on 16/Apr/20

$Is E the focus ?$
Commented by ajfour last updated on 16/Apr/20

$N o, n o t h i n g s p e c i a l .$
	Answered by mr W last updated on 16/Apr/20

	$l e t μ = \frac{b}{a}$ $s a y C (a \cos θ, b \sin θ)$ $s a y C A = D E = s$ $s^{2} = {(1 - \cos θ)}^{2} a^{2} + \sin^{2} θ b^{2}$ $\frac{a \cos θ}{b (1 + \sin θ)} = \frac{b}{a}$ $\Rightarrow \frac{\cos θ}{1 + \sin θ} = μ^{2}$ $l e t t = \tan \frac{θ}{2}$ $\frac{\frac{1 - t^{2}}{1 + t^{2}}}{1 + \frac{2 t}{1 + t^{2}}} = μ^{2}$ $\Rightarrow \frac{1 - t}{1 + t} = μ^{2}$ $\Rightarrow t = \frac{1 - μ^{2}}{1 + μ^{2}}$ $D E = \frac{b}{b (1 + \sin θ)} \times D C$ $s^{2} = \frac{1}{{(1 + \sin θ)}^{2}} \times {D C}^{2} = \frac{b^{2} {(1 + \sin θ)}^{2} + a^{2} \cos^{2} θ}{{(1 + \sin θ)}^{2}}$ $s^{2} = b^{2} + \frac{a^{2} \cos^{2} θ}{{(1 + \sin θ)}^{2}} = b^{2} + (\frac{1 - \sin θ}{1 + \sin θ}) a^{2}$ $b^{2} + (\frac{1 - \sin θ}{1 + \sin θ}) a^{2} = {(1 - \cos θ)}^{2} a^{2} + \sin^{2} θ b^{2}$ $\Rightarrow \frac{1 - \sin θ}{1 + \sin θ} + 2 \cos θ - (1 - μ^{2}) \cos^{2} θ = 1$ $\frac{1 - \frac{2 t}{1 + t^{2}}}{1 + \frac{2 t}{1 + t^{2}}} + 2 \times \frac{1 - t^{2}}{1 + t^{2}} - (1 - μ^{2}) {(\frac{1 - t^{2}}{1 + t^{2}})}^{2} = 1$ ${(\frac{1 - t}{1 + t})}^{2} + (\frac{1 - t^{2}}{1 + t^{2}}) [2 - (1 - μ^{2}) (\frac{1 - t^{2}}{1 + t^{2}})] = 1$ $μ^{4} + (\frac{1 - {(\frac{1 - μ^{2}}{1 + μ^{2}})}^{2}}{1 + {(\frac{1 - μ^{2}}{1 + μ^{2}})}^{2}}) [2 - (1 - μ^{2}) (\frac{1 - {(\frac{1 - μ^{2}}{1 + μ^{2}})}^{2}}{1 + {(\frac{1 - μ^{2}}{1 + μ^{2}})}^{2}})] = 1$ $μ^{4} + (\frac{2 μ^{2}}{1 + μ^{4}}) [2 - (1 - μ^{2}) (\frac{2 μ^{2}}{1 + μ^{4}})] = 1$ $μ^{4} + \frac{4 μ^{2} (1 - μ^{2} + 2 μ^{4})}{{(1 + μ^{4})}^{2}} = 1$ $4 μ^{2} (1 - μ^{2} + 2 μ^{4}) = (1 - μ^{4}) {(1 + μ^{4})}^{2}$ $μ^{12} + μ^{8} + 8 μ^{6} - 5 μ^{4} + 4 μ^{2} - 1 = 0$ $\Rightarrow μ \approx 0.554632$
	Commented by ajfour last updated on 16/Apr/20

	$T h a n k y o u S i r .$ $N i c e A n s w e r : μ ≊ 0.555$
	Commented by mr W last updated on 16/Apr/20

	${y o u}^{'} r e r i g h t, t h a n k s!$
	Answered by ajfour last updated on 16/Apr/20
	$eq. of CD: y=(a/b)x−b x_E =(b^2 /a) , s^2 =b^2 +(b^4 /a^2 ) let C(x,y) ⇒ (x^2 /a^2 )+(1/b^2 )(((ax)/b)−b)^2 =1 x((1/a^2 )+(a^2 /b^4 ))=((2a)/b^2 ) x= 2a(((a^2 b^2 )/(a^4 +b^4 ))) y=(a/b)(2a)(((a^2 b^2 )/(a^4 +b^4 )))−b y=b(((a^4 −b^4 )/(a^4 +b^4 ))) (a−x)^2 +y^2 =s^2 a^2 {1−(((2a^2 b^2 )/(a^4 +b^4 )))}^2 +b^2 (((a^4 −b^4 )/(a^4 +b^4 )))^2 = b^2 (((a^2 +b^2 )/a^2 )) (((a^2 −b^2 )^4 )/((a^4 +b^4 )^2 ))+((b^2 (a^2 −b^2 )^2 (a^2 +b^2 )^2 )/(a^2 (a^4 +b^4 )^2 )) = (b^2 /a^2 )(((a^2 +b^2 )/a^2 )) ⇒ let (b^2 /a^2 )=μ^2 = t t(1+t)(1+t^2 )^2 =(1−t)^4 + t(1−t)^2 (1+t)^2 ⇒ t(1+t)(1+t^2 )^2 =(1−t)^2 ×{(1−t)^2 +t(1+t)^2 } t=μ^2 ≈0.30762 ⇒ μ ≈0.554635$
	$e q . o f C D : y = \frac{a}{b} x - b$ $x_{E} = \frac{b^{2}}{a}, s^{2} = b^{2} + \frac{b^{4}}{a^{2}}$ $l e t C (x, y)$ $\Rightarrow \frac{x^{2}}{a^{2}} + \frac{1}{b^{2}} {(\frac{a x}{b} - b)}^{2} = 1$ $x (\frac{1}{a^{2}} + \frac{a^{2}}{b^{4}}) = \frac{2 a}{b^{2}}$ $x = 2 a (\frac{a^{2} b^{2}}{a^{4} + b^{4}})$ $y = \frac{a}{b} (2 a) (\frac{a^{2} b^{2}}{a^{4} + b^{4}}) - b$ $y = b (\frac{a^{4} - b^{4}}{a^{4} + b^{4}})$ ${(a - x)}^{2} + y^{2} = s^{2}$ $a^{2} {1 - (\frac{2 a^{2} b^{2}}{a^{4} + b^{4}})}^{2} + b^{2} {(\frac{a^{4} - b^{4}}{a^{4} + b^{4}})}^{2}$ $= b^{2} (\frac{a^{2} + b^{2}}{a^{2}})$ $\frac{{(a^{2} - b^{2})}^{4}}{{(a^{4} + b^{4})}^{2}} + \frac{b^{2} {(a^{2} - b^{2})}^{2} {(a^{2} + b^{2})}^{2}}{a^{2} {(a^{4} + b^{4})}^{2}}$ $= \frac{b^{2}}{a^{2}} (\frac{a^{2} + b^{2}}{a^{2}})$ $\Rightarrow l e t \frac{b^{2}}{a^{2}} = μ^{2} = t$ $t (1 + t) {(1 + t^{2})}^{2} = {(1 - t)}^{4} +$ $t {(1 - t)}^{2} {(1 + t)}^{2}$ $\Rightarrow t (1 + t) {(1 + t^{2})}^{2} = {(1 - t)}^{2} \times {{(1 - t)}^{2} + t {(1 + t)}^{2}}$ $t = μ^{2} \approx 0.30762$ $\Rightarrow μ \approx 0.554635$
	Commented by ajfour last updated on 16/Apr/20

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