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Question Number 184135 by ajfour last updated on 03/Jan/23

Commented by mr W last updated on 03/Jan/23

${P Q}_{m i n} = b \sqrt{\frac{1}{2} (1 + \frac{a}{\sqrt{a^{2} + b^{2}}})}$
	Answered by ajfour last updated on 03/Jan/23

	Commented by ajfour last updated on 03/Jan/23
	$tan 2θ=(b/a) PQ^2 =l^2 =t^2 +(b−r)^2 =(r^2 /(tan^2 θ))+(b−r)^2 =(r^2 /m^2 )+(b−r)^2 ((d(l^2 ))/dr)=0 ⇒ b=r(1+(1/m^2 )) l^2 =r^2 ((1/m^2 )+(1/m^4 )) l=(b/( (√(m^2 +1)))) tan 2θ=(b/a)=((2m)/(1−m^2 )) ⇒ m^2 −1+((2am)/b)=0 m=−(a/b)+(√((a^2 /b^2 )+1)) m^2 +1=2−((2am)/b) =2+((2a^2 )/b^2 )−((2a)/b)(√((a^2 /b^2 )+1)) =2{((a^2 +b^2 )/b^2 )−(a/b)(√((a^2 +b^2 )/b^2 ))} l^2 =((b^4 /2)/( (√(a^2 +b^2 ))((√(a^2 +b^2 ))−a))) =(((b^2 /2)(a+(√(a^2 +b^2 ))))/( (√(a^2 +b^2 )))) ⇒ l=(b/( (√2)))(√(1+(a/( (√(a^2 +b^2 ))))))$
	$\tan 2 θ = \frac{b}{a}$ ${P Q}^{2} = l^{2} = t^{2} + {(b - r)}^{2}$ $= \frac{r^{2}}{\tan^{2} θ} + {(b - r)}^{2} = \frac{r^{2}}{m^{2}} + {(b - r)}^{2}$ $\frac{d (l^{2})}{d r} = 0 \Rightarrow b = r (1 + \frac{1}{m^{2}})$ $l^{2} = r^{2} (\frac{1}{m^{2}} + \frac{1}{m^{4}})$ $l = \frac{b}{\sqrt{m^{2} + 1}}$ $\tan 2 θ = \frac{b}{a} = \frac{2 m}{1 - m^{2}}$ $\Rightarrow m^{2} - 1 + \frac{2 a m}{b} = 0$ $m = - \frac{a}{b} + \sqrt{\frac{a^{2}}{b^{2}} + 1}$ $m^{2} + 1 = 2 - \frac{2 a m}{b}$ $= 2 + \frac{2 a^{2}}{b^{2}} - \frac{2 a}{b} \sqrt{\frac{a^{2}}{b^{2}} + 1}$ $= 2 {\frac{a^{2} + b^{2}}{b^{2}} - \frac{a}{b} \sqrt{\frac{a^{2} + b^{2}}{b^{2}}}}$ $l^{2} = \frac{b^{4} / 2}{\sqrt{a^{2} + b^{2}} (\sqrt{a^{2} + b^{2}} - a)}$ $= \frac{(b^{2} / 2) (a + \sqrt{a^{2} + b^{2}})}{\sqrt{a^{2} + b^{2}}}$ $\Rightarrow l = \frac{b}{\sqrt{2}} \sqrt{1 + \frac{a}{\sqrt{a^{2} + b^{2}}}}$
	Commented by ajfour last updated on 03/Jan/23

	$y e s s i r, t h a n k s .$
	Commented by mr W last updated on 03/Jan/23

	$v e r y n i c e q u e s t i o n a n d s o l u t i o n!$
	Answered by mr W last updated on 03/Jan/23

	Commented by mr W last updated on 04/Jan/23

	$a n o t h e r a p p r o a c h$ $l o c u s o f Q i s b i s e c t o r o f a n g l e ∠ N A B .$ $t h e m i n i m u m o f P Q i s t h e d i s t a n c e$ $f r o m P t o t h i s b i s e c t o r : {P Q}^{'} .$ $θ = \frac{β}{2}$ $\cos β = \frac{a}{\sqrt{a^{2} + b^{2}}}$ $\Rightarrow \cos \frac{β}{2} = \sqrt{\frac{1}{2} (\cos β + 1)} = \sqrt{\frac{1}{2} (1 + \frac{a}{\sqrt{a^{2} + b^{2}}})}$ $\Rightarrow {P Q}^{'} = b \cos θ = b \cos \frac{β}{2}$ $= b \sqrt{\frac{1}{2} (1 + \frac{a}{\sqrt{a^{2} + b^{2}}})} = {P Q}_{m i n}$
	Commented by ajfour last updated on 04/Jan/23

	$t h i s i s b e s t a p p r o a c h, t h a n k s s i r .$
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