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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=(√(a^2 −h_1 ^2 ))+(√(b^2 −h_1 ^2 )) m−(√(a^2 −h_1 ^2 ))=(√(b^2 −h_1 ^2 )) m^2 +a^2 −h_1 ^2 −2m(√(a^2 −h_1 ^2 ))=b^2 −h_1 ^2 m^2 +a^2 −b^2 =2m(√(a^2 −h_1 ^2 )) ⇒(√(a^2 −h_1 ^2 ))=((m^2 +a^2 −b^2 )/(2m)) similarly ⇒(√(p^2 −h_2 ^2 ))=((n^2 +p^2 −q^2 )/(2n)) (√(p^2 −h_2 ^2 ))−(√(a^2 −h_1 ^2 ))=((n−m)/2) ((n^2 +p^2 −q^2 )/(2n))−((m^2 +a^2 −b^2 )/(2m))=((n−m)/2) ((p^2 −q^2 )/n)−((a^2 −b^2 )/m)=0 ⇒((a^2 −b^2 )/(p^2 −q^2 ))=(m/n)

$${m}=\sqrt{{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$${m}−\sqrt{{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} }=\sqrt{{b}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$${m}^{\mathrm{2}} +{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} −\mathrm{2}{m}\sqrt{{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} }={b}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{2}{m}\sqrt{{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$$\Rightarrow\sqrt{{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} }=\frac{{m}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{m}} \\ $$$${similarly} \\ $$$$\Rightarrow\sqrt{{p}^{\mathrm{2}} −{h}_{\mathrm{2}} ^{\mathrm{2}} }=\frac{{n}^{\mathrm{2}} +{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{2}{n}} \\ $$$$ \\ $$$$\sqrt{{p}^{\mathrm{2}} −{h}_{\mathrm{2}} ^{\mathrm{2}} }−\sqrt{{a}^{\mathrm{2}} −{h}_{\mathrm{1}} ^{\mathrm{2}} }=\frac{{n}−{m}}{\mathrm{2}} \\ $$$$\frac{{n}^{\mathrm{2}} +{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{2}{n}}−\frac{{m}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{m}}=\frac{{n}−{m}}{\mathrm{2}} \\ $$$$\frac{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }{{n}}−\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{m}}=\mathrm{0} \\ $$$$\Rightarrow\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }=\frac{{m}}{{n}} \\ $$

Commented by mnjuly1970 last updated on 09/Jan/22

very nice solution sir W...thank you so much...

$$\:\:\:\:{very}\:{nice}\:{solution}\:\mathrm{sir}\:\:\mathrm{W}...\mathrm{thank} \\ $$$$\mathrm{you}\:\mathrm{so}\:\mathrm{much}... \\ $$

Commented by Tawa11 last updated on 09/Jan/22

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

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