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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  again

$$\:\:\:\:\:\:\:\:\:\:\:\:{Q}.\mathrm{158749}\:\:{again} \\ $$

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.  ......

$${Let}\:\:\frac{{a}}{\mathrm{sin}\:\theta}={p} \\ $$$${condider}\:{T}\left(−{h},{k}\right)\:{on}\:{it}. \\ $$$$\frac{{k}}{{p}+{a}−{h}}=\mathrm{tan}\:\theta={m} \\ $$$$\left({a}−{h}\right)^{\mathrm{2}} +\left({p}+{a}−{h}\right)^{\mathrm{2}} {m}^{\mathrm{2}} =\left({r}−{a}\right)^{\mathrm{2}} \\ $$$$\left({b}+{h}\right)^{\mathrm{2}} +\left({p}+{a}−{h}\right)^{\mathrm{2}} {m}^{\mathrm{2}} =\left({r}−{b}\right)^{\mathrm{2}} \\ $$$${subtracting} \\ $$$$\left({a}−{b}−\mathrm{2}{h}\right)\left({a}+{b}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{2}{r}−{a}−{b}\right)\left({b}−{a}\right) \\ $$$$\mathrm{2}{r}={a}+{b}+\left(\frac{{a}+{b}}{{a}−{b}}\right)\left\{\mathrm{2}{h}−\left({a}−{b}\right)\right\} \\ $$$$\frac{{d}\left(\mathrm{2}{r}\right)}{{dh}}=\mathrm{2}\left(\frac{{a}+{b}}{{a}−{b}}\right)\:\:\:\left[{constant}>\mathrm{0}\right] \\ $$$${but}\:{limits}\:{of}\:\boldsymbol{{h}}\:{is}\:{difficult}\:{to} \\ $$$${guess}. \\ $$$$...... \\ $$

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