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Question Number 215828 by CrispyXYZ last updated on 19/Jan/25

Let Γ be a hyperbola with foci F_1 and F_2 , eccentricity e. M is an arbitrary point on Γ. Let x=∠MF_1 F_2 , y=∠MF_2 F_1 Prove that ((∣ cos x − cos y ∣)/(1 − cos x cos y)) = ((2e)/(e^2 +1)).

LetΓbeahyperbolawithfociF1andF2,eccentricitye.MisanarbitrarypointonΓ.Letx=∠MF1F2,y=∠MF2F1Provethat∣cosx−cosy∣1−cosxcosy=2ee2+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)) ✓

sayMF1=p,MF2=qaccordingtodefinitionofhyperbolap−q=constant=2asayF1F2=2cc=eawithe=eccentricityx=α1,y=α2cosα1=p2+4c2−q24cpcosα2=q2+4c2−p24cqcosα1−cosα2=14c(p2+4c2−q2p−q2+4c2−p2q)=14c(p−q+4c2−q2p−4c2−p2q)=14c[p−q+p3−q2−4c2(p−q)pq]=a2c[1+p2+pq+q2−4c2pq]=a2c[1+(p−q)2+3pq−4c2pq]=2ac(1+a2−c2pq)=2e[1+a2(1−e2)pq]1−cosα1cosα2=1−p2+4c2−q24cp×q2+4c2−p24cq=1+(p2−q2+4c2)(p2−q2−4c2)16c2pq=1+(p2−q2)2−16c416c2pq=1+a2(p+q)2−4c44c2pq=1+a2[(p−q)2+4pq]−4c44c2pq=1+a2(a2+pq)−c4c2pq=1+1e2+a2(1−e4)e2pq=e2+1e2[1+a2(1−e2)pq]∣cosα1−cosα2∣1−cosα1cosα2=2e×e2e2+1=2ee2+1✓

Commented by Tawa11 last updated on 19/Jan/25

Weldone sir. Please help me see to Q215845

Weldonesir.PleasehelpmeseetoQ215845

Commented by CrispyXYZ last updated on 20/Jan/25

Great! But I was wondering whether there exists a better solution.

Great!ButIwaswonderingwhetherthereexistsabettersolution.

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.

itdependsonwhatyouunderstandwith‘‘better″inmathematics.ionlyusedthedefinitionofhyperbolaandthedefinitionofeccentricity,soinmyopinionitisthebestway.certainlyonecanalsosolveusingtheequationofhyperbola.butit′sonlyanotherway,notnecessarilyabetterway,sinceoneshouldproveatfirsttheequationofthehyperbola.

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