Question Number 10414 by amir last updated on 07/Feb/17

Commented bymrW1 last updated on 08/Feb/17

the angle between tangent lines is  not a constant!

Commented byamir last updated on 08/Feb/17

hello dear mrW1.thank you so much  for solving this problems.  yes .you are right. angle is not constant  and i shuld fix it.

Answered by mrW1 last updated on 08/Feb/17

Q1:  y=(1/x)  y′=−(1/x^2 )  tangent line at point (t,(1/t)) on curve  is  y−(1/t)=−(1/t^2 )(x−t)  since point (a,−a) lies on tangent line  −a−(1/t)=−(1/t^2 )(a−t)  at^2 +2t−a=0  t_(1,2) =((−2±(√(4+4a^2 )))/(2a))=((−1±(√(1+a^2 )))/a), a≠0  i.e. from point A(a,−a) with a≠0  there are 2 tangent lines:  AD and AE  D(t_1 ,(1/t_1 )) and E(t_2 ,(1/t_2 ))  slope of AD=β_1   slope of AE=β_2   angle between tangent lines=β=β_2 −β_1   tan β_1 =−(1/t_1 ^2 )=−(1/((((−1+(√(1+a^2 )))/a))^2 ))=−(a^2 /(2+a^2 −2(√(1+a^2 ))))  tan β_2 =−(1/t_2 ^2 )=−(1/((((−1−(√(1+a^2 )))/a))^2 ))=−(a^2 /(2+a^2 +2(√(1+a^2 ))))  tan β=((−(a^2 /(2+a^2 +2(√(1+a^2 ))))+(a^2 /(2+a^2 −2(√(1+a^2 )))))/(1+(a^2 /(2+a^2 +2(√(1+a^2 ))))×(a^2 /(2+a^2 −2(√(1+a^2 ))))))  =((a^2 (−2−a^2 +2(√(1+a^2 ))+2+a^2 +2(√(1+a^2 ))))/((2+a^2 +2(√(1+a^2 )))(2+a^2 −2(√(1+a^2 )))+a^4 ))  =((4a^2 (√(1+a^2 )))/((2+a^2 )^2 −4(1+a^2 )+a^4 ))  =((4a^2 (√(1+a^2 )))/(4+4a^2 +a^4 −4(1+a^2 )+a^4 ))  =((2(√(1+a^2 )))/a^2 )  ⇒β=tan^(−1) (((2(√(1+a^2 )))/a^2 ))≠constant    Q2:  perpendicular line at point (t,(1/t))  on curve xy=1 is  y−(1/t)=t^2 (x−t)  since point A(a,−a) lies on the line  −a−(1/t)=t^2 (a−t)  t^4 −at^3 −at−1=0  (t^2 +1)(t^2 −at−1)=0  since t^2 +1≠0  ⇒t^2 −at−1=0  t_(1,2) =((a±(√(a^2 +4)))/2)  i.e. from point A(a,−a) there are  2 perpendicular lines:  AB and AC with  B(t_1 ,(1/t_1 )) and C(t_2 ,(1/t_2 ))  slope of AB=α_1   slope of AC=α_2   angle between them =α=α_2 −α_1   tan α_1 =t_1 ^2 =(((a+(√(a^2 +4)))/2))^2 =((a^2 +2+a(√(a^2 +4)))/2)  tan α_2 =t_2 ^2 =(((a−(√(a^2 +4)))/2))^2 =((a^2 +2−a(√(a^2 +4)))/2)  tan α=(((((a−(√(a^2 +4)))/2))^2 −(((a+(√(a^2 +4)))/2))^2 )/(1+(((a+(√(a^2 +4)))/2))^2 (((a−(√(a^2 +4)))/2))^2 ))  =4×(((a−(√(a^2 +4)))^2 −(a+(√(a^2 +4)))^2 )/(16+(a+(√(a^2 +4)))^2 (a−(√(a^2 +4)))^2 ))  =4×((−2a×2(√(a^2 +4)))/(16+(a^2 −a^2 −4)^2 ))  =−((a(√(a^2 +4)))/2)  ⇒α=tan^(−1) (−((a(√(a^2 +4)))/2))    angle between tangent line and  perpendicular line=γ=β_1 −α_1   tan γ=((−(a^2 /(2+a^2 −2(√(1+a^2 ))))−((a^2 +2+a(√(a^2 +4)))/2))/(1+(a^2 /(2+a^2 −2(√(1+a^2 ))))×((a^2 +2+a(√(a^2 +4)))/2)))  =−((2a^2 +(2+a^2 −2(√(1+a^2 )))(a^2 +2+a(√(a^2 +4))))/(2(2+a^2 −2(√(1+a^2 )))+a^2 (a^2 +2+a(√(a^2 +4)))))   =−((2a^2 +(2+a^2 −2(√(1+a^2 )))(a^2 +2+a(√(a^2 +4))))/(4(1+a^2 )−4(√(1+a^2 ))+a^4 +a^3 (√(a^2 +4))))   ⇒γ=tan^(−1) (−((2a^2 +(2+a^2 −2(√(1+a^2 )))(a^2 +2+a(√(a^2 +4))))/(4(1+a^2 )−4(√(1+a^2 ))+a^4 +a^3 (√(a^2 +4)))))

Commented bymrW1 last updated on 08/Feb/17

Answered by mrW1 last updated on 08/Feb/17

Q3:  x_B =((a+(√(a^2 +4)))/2)  y_B =(1/x_B )=((−a+(√(a^2 +4)))/2)  x_D =((−1+(√(1+a^2 )))/a)  y_D =(1/x_D )=((1+(√(1+a^2 )))/a)    area F=F_1 +F_2 −F_3 −F_4   F_1 =a(x_B −x_D )=a(((a+(√(a^2 +4)))/2)−((−1+(√(1+a^2 )))/a))=  =((a^2 +a(√(a^2 +4))+2−2(√(1+a^2 )))/2)  F_2 =∫_x_D  ^x_B  (1/x)dx=(ln x)∣_x_D  ^x_B  =ln (x_B /x_D )=ln (((a+(√(a^2 +4)))/2)/((−1+(√(1+a^2 )))/a))  =ln ∣((a(a+(√(a^2 +4))))/(2((√(1+a^2 ))−1)))∣  F_3 =(1/2)(x_B −a)(a+y_B )=(1/2)(((a+(√(a^2 +4)))/2)−a)(a+((−a+(√(a^2 +4)))/2))=((((√(a^2 +4))−a)((√(a^2 +4))+a))/8)  =(1/2)  F_4 =(1/2)(a−x_D )(a+y_D )=(1/2)(a−((−1+(√(1+a^2 )))/a))(a+((1+(√(1+a^2 )))/a))=(((a^2 +1−(√(1+a^2 )))(a^2 +1+(√(1+a^2 ))))/(2a^2 ))  =((1+a^2 )/2)  F=((a(√(a^2 +4))−2(√(1+a^2 )))/2)+ln ∣((a(a+(√(a^2 +4))))/(2((√(1+a^2 ))−1)))∣