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Question Number 158803 by ajfour last updated on 09/Nov/21

Commented by ajfour last updated on 09/Nov/21

$Q . 158749 a g a i n$
	Answered by ajfour last updated on 09/Nov/21

	Commented by ajfour last updated on 09/Nov/21
	$Let (a/(sin θ))=p condider T(−h,k) on it. (k/(p+a−h))=tan θ=m (a−h)^2 +(p+a−h)^2 m^2 =(r−a)^2 (b+h)^2 +(p+a−h)^2 m^2 =(r−b)^2 subtracting (a−b−2h)(a+b) =(2r−a−b)(b−a) 2r=a+b+(((a+b)/(a−b))){2h−(a−b)} ((d(2r))/dh)=2(((a+b)/(a−b))) [constant>0] but limits of h is difficult to guess. ......$
	$L e t \frac{a}{\sin θ} = p$ $c o n d i d e r T (- h, k) o n i t .$ $\frac{k}{p + a - h} = \tan θ = m$ ${(a - h)}^{2} + {(p + a - h)}^{2} m^{2} = {(r - a)}^{2}$ ${(b + h)}^{2} + {(p + a - h)}^{2} m^{2} = {(r - b)}^{2}$ $s u b t r a c t i n g$ $(a - b - 2 h) (a + b)$ $= (2 r - a - b) (b - a)$ $2 r = a + b + (\frac{a + b}{a - b}) {2 h - (a - b)}$ $\frac{d (2 r)}{d h} = 2 (\frac{a + b}{a - b}) [c o n s t a n t > 0]$ $b u t l i m i t s o f h i s d i f f i c u l t t o$ $g u e s s .$ $. . . . . .$
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