Question Number 103147 by ajfour last updated on 13/Jul/20

Answered by mr W last updated on 13/Jul/20

A(a,0)  B(0,(√(p^2 −a^2 )))  C(0,c)  D((√(q^2 −c^2 )),0)  a(√(p^2 −a^2 ))=c(√(q^2 −c^2 ))   ...(i)  intersection AB and CD at (h,k)  (h/a)+(k/(√(p^2 −a^2 )))=1  (h/(√(q^2 −c^2 )))+(k/c)=1  ((1/(ac))−(1/(√((p^2 −a^2 )(q^2 −c^2 )))))h=(1/c)−(1/(√(p^2 −a^2 )))  ⇒h=(((1/c)−(1/(√(p^2 −a^2 ))))/((1/(ac))−(1/(√((p^2 −a^2 )(q^2 −c^2 ))))))  (c−(√(p^2 −a^2 )))h=(1/2)c(√(q^2 −c^2 ))  ⇒(c−(√(p^2 −a^2 )))(((1/c)−(1/(√(p^2 −a^2 ))))/((1/(ac))−(1/(√((p^2 −a^2 )(q^2 −c^2 ))))))=(1/2)c(√(q^2 −c^2 ))   ...(ii)  from (i):  c^4 −q^2 c^2 +a^2 (p^2 −a^2 )=0  c^2 =((q^2 +(√(q^4 −4a^2 (p^2 −a^2 ))))/2)  c=(√((q^2 +(√(q^4 −4a^2 (p^2 −a^2 ))))/2))  put this into (ii) to find a...

Commented bymr W last updated on 13/Jul/20

Commented byajfour last updated on 13/Jul/20

Thanks Sir, but i think, i am able  to find the exact expression,  kindly check the same..

Answered by ajfour last updated on 13/Jul/20

let ∠OAB=θ  ;  ∠OCD=φ  A_1 =A_2 =A_3 =△  4△=p^2 sin θcos θ=q^2 sin φcos φ                              ......(i)  Let  P  be point of intersection of  AB and CD.  Also  say  AP = r  ,  CP = s  OD=qsin φ ,   OB=psin θ  2△=(pcos θ−qsin φ)(rsin θ)  ..(ii)  2△=(qcos φ−psin θ)(ssin φ)  ..(iii)  2△=(qsin φ)(rsin θ)                    +(psin θ)(ssin φ)    ...(iv)  using (ii), (iii) in (iv)  ((qsin φ)/(pcos θ−qsin φ))+((psin θ)/(qcos φ−psin θ))=1  ⇒  (1/((((pcos θ)/(qsin φ)))−1))+(1/((((qcos φ)/(psin θ)))−1))=1                                                          And from (i)            ((qcos φ)/(psin θ))=((pcos θ)/(qsin φ)) = R      ....(I)  So     (1/(R−1))+(1/(R−1))=1  ⇒     R−1=2    ⇒   R=3  hence   q cos φ=3psin θ  qsin φ=(√(q^2 −9p^2 sin ^2 θ))  substituting this in (I):   ⇒   ((pcos θ)/(√(q^2 −9p^2 sin ^2 θ)))=3  ⇒  p^2 cos ^2 θ=9q^2 −81p^2 (1−cos ^2 θ)  ⇒   p^2 cos ^2 θ = ((81p^2 −9q^2 )/(80))  &     q^2 cos ^2 φ=9(p^2 −p^2 cos ^2 θ)                              = 9(p^2 −((81p^2 −9q^2 )/(80)))  ⇒    q^2 cos ^2 φ = ((81q^2 −9p^2 )/(80))  AC^( 2) = p^2 cos ^2 θ+q^2 cos ^2 φ  ⇒  AC ^2 =(9/(10))(p^2 +q^2 )         AC = 3(√((p^2 +q^2 )/(10))) .  eg.  p=5 , q=4  AC=3(√((25+16)/(10))) = 3(√((41)/(10)))    pcos θ=(3/4)(√((9p^2 −q^2 )/5))                  = (3/4)(√((225−16)/5)) = (3/4)(√((209)/5))    psin θ=(√(25−((9×209)/(80))))                 = (1/4)(√((119)/5))    qcos φ = (3/4)(√((9q^2 −p^2 )/5))             = (3/4)(√((144−25)/5)) =(3/4)(√((119)/5))   qsin φ = (√(16−((9×119)/(80)))) =(1/4)(√((209)/5))  I shall try to graph it...

Commented byajfour last updated on 13/Jul/20

Commented bymr W last updated on 13/Jul/20

fantastically solved!