Question Number 163656 by mnjuly1970 last updated on 09/Jan/22
Answered by mr W last updated on 09/Jan/22
Commented by mr W last updated on 09/Jan/22
$${m}=\sqrt{{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$${m}−\sqrt{{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} }=\sqrt{{b}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$${m}^{\mathrm{2}} +{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}{m}\sqrt{{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} }={b}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{2}{m}\sqrt{{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$$\Rightarrow\sqrt{{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} }=\frac{{m}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{m}} \\ $$$${similarly} \\ $$$$\Rightarrow\sqrt{{p}^{\mathrm{2}} −{h}_{\mathrm{2}} ^{\mathrm{2}} }=\frac{{n}^{\mathrm{2}} +{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{2}{n}} \\ $$$$ \\ $$$$\sqrt{{p}^{\mathrm{2}} −{h}_{\mathrm{2}} ^{\mathrm{2}} }−\sqrt{{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} }=\frac{{n}−{m}}{\mathrm{2}} \\ $$$$\frac{{n}^{\mathrm{2}} +{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{2}{n}}−\frac{{m}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{m}}=\frac{{n}−{m}}{\mathrm{2}} \\ $$$$\frac{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{{n}}−\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{m}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }=\frac{{m}}{{n}} \\ $$
Commented by mnjuly1970 last updated on 09/Jan/22
$$\:\:\:\:{very}\:{nice}\:{solution}\:\mathrm{sir}\:\:\mathrm{W}…\mathrm{thank} \\ $$$$\mathrm{you}\:\mathrm{so}\:\mathrm{much}… \\ $$
Commented by Tawa11 last updated on 09/Jan/22
$$\mathrm{Great}\:\mathrm{sir}. \\ $$