Question Number 215828 by CrispyXYZ last updated on 19/Jan/25
$$\mathrm{Let}\:\Gamma\:\mathrm{be}\:\mathrm{a}\:\mathrm{hyperbola}\:\mathrm{with}\:\mathrm{foci}\:{F}_{\mathrm{1}} \:\mathrm{and}\:{F}_{\mathrm{2}} ,\: \\ $$$$\mathrm{eccentricity}\:{e}.\:{M}\:\mathrm{is}\:\mathrm{an}\:\mathrm{arbitrary}\:\mathrm{point}\:\mathrm{on}\:\Gamma. \\ $$$$\mathrm{Let}\:{x}=\angle{MF}_{\mathrm{1}} {F}_{\mathrm{2}} ,\:{y}=\angle{MF}_{\mathrm{2}} {F}_{\mathrm{1}} \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\frac{\mid\:\mathrm{cos}\:{x}\:−\:\mathrm{cos}\:{y}\:\mid}{\mathrm{1}\:−\:\mathrm{cos}\:{x}\:\mathrm{cos}\:{y}}\:=\:\frac{\mathrm{2}{e}}{{e}^{\mathrm{2}} +\mathrm{1}}. \\ $$
Answered by mr W last updated on 19/Jan/25
Commented by mr W last updated on 19/Jan/25
![say MF_1 =p, MF_2 =q according to definition of hyperbola p−q=constant=2a say F_1 F_2 =2c c=ea with e=eccentricity x=α_1 , y=α_2 cos α_1 =((p^2 +4c^2 −q^2 )/(4cp)) cos α_2 =((q^2 +4c^2 −p^2 )/(4cq)) cos α_1 −cos α_2 =(1/(4c))(((p^2 +4c^2 −q^2 )/p)−((q^2 +4c^2 −p^2 )/q)) =(1/(4c))(p−q+((4c^2 −q^2 )/p)−((4c^2 −p^2 )/q)) =(1/(4c))[p−q+((p^3 −q^2 −4c^2 (p−q))/(pq))] =(a/(2c))[1+((p^2 +pq+q^2 −4c^2 )/(pq))] =(a/(2c))[1+(((p−q)^2 +3pq−4c^2 )/(pq))] =((2a)/c)(1+((a^2 −c^2 )/(pq))) =(2/e)[1+((a^2 (1−e^2 ))/(pq))] 1−cos α_1 cos α_2 =1−((p^2 +4c^2 −q^2 )/(4cp))×((q^2 +4c^2 −p^2 )/(4cq)) =1+(((p^2 −q^2 +4c^2 )(p^2 −q^2 −4c^2 ))/(16c^2 pq)) =1+(((p^2 −q^2 )^2 −16c^4 )/(16c^2 pq)) =1+((a^2 (p+q)^2 −4c^4 )/(4c^2 pq)) =1+((a^2 [(p−q)^2 +4pq]−4c^4 )/(4c^2 pq)) =1+((a^2 (a^2 +pq)−c^4 )/(c^2 pq)) =1+(1/e^2 )+((a^2 (1−e^4 ))/(e^2 pq)) =((e^2 +1)/e^2 )[1+((a^2 (1−e^2 ))/(pq))] ((∣cos α_1 −cos α_2 ∣)/(1−cos α_1 cos α_2 ))=(2/e)×(e^2 /(e^2 +1))=((2e)/(e^2 +1)) ✓](https://www.tinkutara.com/question/Q215847.png)
$${say}\:{MF}_{\mathrm{1}} ={p},\:{MF}_{\mathrm{2}} ={q} \\ $$$${according}\:{to}\:{definition}\:{of}\:{hyperbola} \\ $$$${p}−{q}={constant}=\mathrm{2}{a} \\ $$$${say}\:{F}_{\mathrm{1}} {F}_{\mathrm{2}} =\mathrm{2}{c} \\ $$$${c}={ea}\:{with}\:{e}={eccentricity} \\ $$$${x}=\alpha_{\mathrm{1}} ,\:{y}=\alpha_{\mathrm{2}} \\ $$$$\mathrm{cos}\:\alpha_{\mathrm{1}} =\frac{{p}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{4}{cp}} \\ $$$$\mathrm{cos}\:\alpha_{\mathrm{2}} =\frac{{q}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{4}{cq}} \\ $$$$\mathrm{cos}\:\alpha_{\mathrm{1}} −\mathrm{cos}\:\alpha_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}{c}}\left(\frac{{p}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} −{q}^{\mathrm{2}} }{{p}}−\frac{{q}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} −{p}^{\mathrm{2}} }{{q}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}{c}}\left({p}−{q}+\frac{\mathrm{4}{c}^{\mathrm{2}} −{q}^{\mathrm{2}} }{{p}}−\frac{\mathrm{4}{c}^{\mathrm{2}} −{p}^{\mathrm{2}} }{{q}}\right) \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}{c}}\left[{p}−{q}+\frac{{p}^{\mathrm{3}} −{q}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} \left({p}−{q}\right)}{{pq}}\right] \\ $$$$\:\:\:\:=\frac{{a}}{\mathrm{2}{c}}\left[\mathrm{1}+\frac{{p}^{\mathrm{2}} +{pq}+{q}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} }{{pq}}\right] \\ $$$$\:\:\:\:=\frac{{a}}{\mathrm{2}{c}}\left[\mathrm{1}+\frac{\left({p}−{q}\right)^{\mathrm{2}} +\mathrm{3}{pq}−\mathrm{4}{c}^{\mathrm{2}} }{{pq}}\right] \\ $$$$\:\:\:=\frac{\mathrm{2}{a}}{{c}}\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{pq}}\right) \\ $$$$\:\:\:=\frac{\mathrm{2}}{{e}}\left[\mathrm{1}+\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right)}{{pq}}\right] \\ $$$$\mathrm{1}−\mathrm{cos}\:\alpha_{\mathrm{1}} \mathrm{cos}\:\alpha_{\mathrm{2}} =\mathrm{1}−\frac{{p}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} −{q}^{\mathrm{2}} }{\mathrm{4}{cp}}×\frac{{q}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} −{p}^{\mathrm{2}} }{\mathrm{4}{cq}} \\ $$$$\:\:\:=\mathrm{1}+\frac{\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} \right)\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} \right)}{\mathrm{16}{c}^{\mathrm{2}} {pq}} \\ $$$$\:\:\:=\mathrm{1}+\frac{\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{16}{c}^{\mathrm{4}} }{\mathrm{16}{c}^{\mathrm{2}} {pq}} \\ $$$$\:\:\:=\mathrm{1}+\frac{{a}^{\mathrm{2}} \left({p}+{q}\right)^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{4}} }{\mathrm{4}{c}^{\mathrm{2}} {pq}} \\ $$$$\:\:\:=\mathrm{1}+\frac{{a}^{\mathrm{2}} \left[\left({p}−{q}\right)^{\mathrm{2}} +\mathrm{4}{pq}\right]−\mathrm{4}{c}^{\mathrm{4}} }{\mathrm{4}{c}^{\mathrm{2}} {pq}} \\ $$$$\:\:\:=\mathrm{1}+\frac{{a}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{pq}\right)−{c}^{\mathrm{4}} }{{c}^{\mathrm{2}} {pq}} \\ $$$$\:\:\:=\mathrm{1}+\frac{\mathrm{1}}{{e}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{4}} \right)}{{e}^{\mathrm{2}} {pq}} \\ $$$$\:\:\:=\frac{{e}^{\mathrm{2}} +\mathrm{1}}{{e}^{\mathrm{2}} }\left[\mathrm{1}+\frac{{a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right)}{{pq}}\right] \\ $$$$\frac{\mid\mathrm{cos}\:\alpha_{\mathrm{1}} −\mathrm{cos}\:\alpha_{\mathrm{2}} \mid}{\mathrm{1}−\mathrm{cos}\:\alpha_{\mathrm{1}} \mathrm{cos}\:\alpha_{\mathrm{2}} }=\frac{\mathrm{2}}{{e}}×\frac{{e}^{\mathrm{2}} }{{e}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{2}{e}}{{e}^{\mathrm{2}} +\mathrm{1}}\:\checkmark \\ $$
Commented by Tawa11 last updated on 19/Jan/25
$$\mathrm{Weldone}\:\mathrm{sir}. \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{see}\:\mathrm{to}\:\:\mathrm{Q215845} \\ $$
Commented by CrispyXYZ last updated on 20/Jan/25
$$\mathrm{Great}!\:\mathrm{But}\:\mathrm{I}\:\mathrm{was}\:\mathrm{wondering}\:\mathrm{whether}\:\mathrm{there}\: \\ $$$$\mathrm{exists}\:\mathrm{a}\:\mathrm{better}\:\mathrm{solution}. \\ $$
Commented by mr W last updated on 20/Jan/25
$${it}\:{depends}\:{on}\:{what}\:{you}\:{understand} \\ $$$${with}\:“{better}''\:{in}\:{mathematics}.\:{i}\:{only} \\ $$$${used}\:{the}\:{definition}\:{of}\:{hyperbola}\:{and} \\ $$$${the}\:{definition}\:{of}\:{eccentricity},\:{so}\:{in} \\ $$$${my}\:{opinion}\:{it}\:{is}\:{the}\:{best}\:{way}. \\ $$$${certainly}\:{one}\:{can}\:{also}\:{solve}\:{using} \\ $$$${the}\:{equation}\:{of}\:{hyperbola}.\:{but}\:{it}'{s} \\ $$$${only}\:{an}\:{other}\:{way},\:{not}\:{necessarily} \\ $$$${a}\:{better}\:{way},\:{since}\:{one}\:{should}\:{prove} \\ $$$${at}\:{first}\:{the}\:{equation}\:{of}\:{the}\:{hyperbola}. \\ $$