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Question Number 124627 by ajfour last updated on 04/Dec/20

Commented by ajfour last updated on 04/Dec/20

$p a r a b o l a i s y = x^{2}, r a d i u s = 1 .$ $F i n d {A B}_{m i n} .$
	Answered by mr W last updated on 05/Dec/20

	Commented by mr W last updated on 05/Dec/20

	Commented by mr W last updated on 05/Dec/20
	$C(h,1) x_D =h−sin θ y_D =1+cos θ 1+cos θ=(h−sin θ)^2 ...(i) tan θ=2(h−sin θ) ...(ii) ⇒4(1+cos θ)=tan^2 θ=(1/(cos^2 θ))−1 ⇒(1/(cos^3 θ))−(5/(cos θ))−4=0 ⇒((1/(cos θ))+1)((1/(cos^2 θ))−(1/(cos θ))−4)=0 ⇒(1/(cos θ))=((1+(√(17)))/2)=(8/( (√(17))−1)) ⇒θ=cos^(−1) (((√(17))−1)/8)≈67.021343° ⇒h=sin θ+((tan θ)/2)≈2.099798 x_E =h+sin ϕ y_E =1+cos ϕ eqn. of AB: y=1+cos ϕ−tan ϕ (x−h−sin ϕ) 0=1+cos ϕ−tan ϕ (x_B −h−sin ϕ) ⇒x_B =h+sin ϕ+((1+cos ϕ)/(tan ϕ)) x_A ^2 =1+cos ϕ−tan ϕ (x_A −h−sin ϕ) x_A ^2 +tan ϕ x_A −[1+cos ϕ+tan ϕ (h+sin ϕ)]=0 ⇒x_A =(1/2)(−tan ϕ+(√(tan^2 ϕ+4[1+cos ϕ+tan ϕ (h+sin ϕ)])) s=AB=((x_B −x_A )/(cos ϕ)) s=(1/(cos ϕ)){h+sin ϕ+((1+cos ϕ)/(tan ϕ))+((tan ϕ)/2)−(√(((tan^2 ϕ)/4)+1+cos ϕ+tan ϕ (h+sin ϕ)))} ⇒s_(min) ≈3.8919 at ϕ=53.4583°$
	$C (h, 1)$ $x_{D} = h - \sin θ$ $y_{D} = 1 + \cos θ$ $1 + \cos θ = {(h - \sin θ)}^{2} . . . (i)$ $\tan θ = 2 (h - \sin θ) . . . (i i)$ $\Rightarrow 4 (1 + \cos θ) = \tan^{2} θ = \frac{1}{\cos^{2} θ} - 1$ $\Rightarrow \frac{1}{\cos^{3} θ} - \frac{5}{\cos θ} - 4 = 0$ $\Rightarrow (\frac{1}{\cos θ} + 1) (\frac{1}{\cos^{2} θ} - \frac{1}{\cos θ} - 4) = 0$ $\Rightarrow \frac{1}{\cos θ} = \frac{1 + \sqrt{17}}{2} = \frac{8}{\sqrt{17} - 1}$ $\Rightarrow θ = \cos^{- 1} \frac{\sqrt{17} - 1}{8} \approx 67.021343 °$ $\Rightarrow h = \sin θ + \frac{\tan θ}{2} \approx 2.099798$ $x_{E} = h + \sin φ$ $y_{E} = 1 + \cos φ$ $e q n . o f A B :$ $y = 1 + \cos φ - \tan φ (x - h - \sin φ)$ $0 = 1 + \cos φ - \tan φ (x_{B} - h - \sin φ)$ $\Rightarrow x_{B} = h + \sin φ + \frac{1 + \cos φ}{\tan φ}$ $x_{A}^{2} = 1 + \cos φ - \tan φ (x_{A} - h - \sin φ)$ $x_{A}^{2} + \tan φ x_{A} - [1 + \cos φ + \tan φ (h + \sin φ)] = 0$ $\Rightarrow x_{A} = \frac{1}{2} (- \tan φ + \sqrt{\tan^{2} φ + 4 [1 + \cos φ + \tan φ (h + \sin φ)]}$ $s = A B = \frac{x_{B} - x_{A}}{\cos φ}$ $s = \frac{1}{\cos φ} {h + \sin φ + \frac{1 + \cos φ}{\tan φ} + \frac{\tan φ}{2} - \sqrt{\frac{\tan^{2} φ}{4} + 1 + \cos φ + \tan φ (h + \sin φ)}}$ $\Rightarrow s_{m i n} \approx 3.8919 a t φ = 53.4583 °$
	Commented by mr W last updated on 05/Dec/20

	Commented by ajfour last updated on 05/Dec/20

	$U n d e r s t o o d y o u r s o l u t i o n, S i r!$ $T h a n k s a l o t .$
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