Question Number 10223 by amir last updated on 30/Jan/17

Answered by mrW1 last updated on 03/Feb/17

to question 1:    xy=1  y=(1/x)  y′=−(1/x^2 )  let B a point on the curve and AB is  a orthogonal line to the curve.  let point B have the coordinates (t, s).  s=(1/t)    the slope of tangent line on point B:  m_t =y′(t)=−(1/t^2 )    the slope of the orthogonal line:  m_o =−(1/m_t )=t^2     the equation of the orthogonal line:  y−s=m_o (x−t)  y−(1/t)=t^2 (x−t)    since A(a,a) is a point on it, we get  a−(1/t)=t^2 (a−t)  t^4 −at^3 +at−1=0  (t+1)(t−1)(t^2 −at+1)=0  ⇒t_1 =−1 (point D) or  ⇒t_2 =1 (point C) or  ⇒t−at+1=0  ⇒⇒t_(3,4) =((a±(√(a^2 −4)))/2)  if ∣a∣<2, t_(3,4)  =unreal  if ∣a∣=2, t_(3,4)  =1=t_2   if ∣a∣>2, t_(3,4)  =((a±(√(a^2 −4)))/2) (point B and B′)    i.e. for any point A(a,a) there are  at least 2 orthogonal lines (AC, AD).  if ∣a∣>2, there are even 2 additional  orthogonal lines (AB,AB′).

Commented bymrW1 last updated on 02/Feb/17

Commented byamir last updated on 02/Feb/17

thank you so much.but Q no.2 and no.3 ?

Commented byamir last updated on 02/Feb/17

with which program i can draw this    diagrams?

Commented bymrW1 last updated on 03/Feb/17

please try the app geogebra.

Answered by mrW1 last updated on 04/Feb/17

to question 2:    if ∣a∣≤2, i.e. −2≤a≤2, there are only  2 orthogonal lines from point A:  AD with slope = 45°  AC with slope = 45°  the angel between them is 0°.    if ∣a∣>2, i.e. a<−2 or a>2, there are  4 orthogonal lines from point A:  AD with slope = 45°  AC with slope = 45°  the angel between them is 0°.  AB with slope angel, lets say α_1   AB′ with slope angel, lets say α_2   the angel between them is, lets say α  α=α_1 −α_2   tan α_1 =−(1/(f′(t_3 )))=t_3 ^2 =(((a+(√(a^2 −4)))/2))^2   tan α_2 =−(1/(f′(t_4 )))=t_4 ^2 =(((a−(√(a^2 −4)))/2))^2   tan α=tan (α_1 −α_2 )=((tan α_1 −tan α_2 )/(1+tan _1 tan α_2 ))  =(((((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×(((a+(√(a^2 −4))+a−(√(a^2 −4)))(a+(√(a^2 −4))−a+(√(a^2 −4))))/(16+(a^2 −a^2 +4)^2 ))  =4×(((2a)(2(√(a^2 −4))))/(32))  =((a(√(a^2 −4)))/2)  α=tan^(−1) (((a(√(a^2 −4)))/2))

Commented byamir last updated on 04/Feb/17

thank you for evrey thing.but please  check your calculation.i got the angle  α=tan^(−1) ((1/2)a(√(a^2 −4)))

Commented bymrW1 last updated on 04/Feb/17

your are right. it is fixed now. thank you!

Commented bymrW1 last updated on 04/Feb/17

what about Q1 and Q2 if we have point A(a,−a)?

Commented byamir last updated on 08/Feb/17

in case of point B(a,−a) ,there are  only 2 perpendiculare lines.  y=(1/x)⇒y^′ =−(1/x^2 )⇒m_t =−(1/t^2 ),m_p =+t^2   that M(t,(1/t))  and N(s,(1/s)) are the intersection points  of the curve and the per.lines.  olso m_p  be the slope of per_ .line  y−y_M =m_p (x−x_M )⇒y−(1/t)=t^2 (x−t)  −a−(1/t)=t^2 (a−t)⇒t^4 −at^3 −at−1=0  (t^2 +1)(t^2 −at−1)=0, t^2 +1≠0  ⇒t=((a±(√(a^2 +4)))/2)⇒ (((M(((a+(√(a^2 +4)))/2),((a−(√(a^2 +4)))/(−2))))),((N(((a−(√(a^2 +4)))/2),((a+(√(a^2 +4)))/(−2))))) )  m_(BN) =((−a−((a+(√(a^2 +4)))/(−2)))/(a−((a−(√(a^2 +4)))/2)))=(1/4)(a−(√(a^2 +4)))^2   m_(BM) =((−a−((a−(√(a^2 +4)))/(−2)))/(a−((a+(√(a^2 +4)))/2)))=(1/4)(a+(√(a^2 +4)))^2   tan α=∣((m_(BN) −m_(BM) )/(1+m_(BN) ×m_(BM) ))∣=  ∣(((1/4)(a−(√(a^2 +4)))^2 −(1/4)(a+(√(a^2 +4)))^2 )/(1+(1/4)(a−(√(a^2 +4)))^2 ×(1/4)(a+(√(a^2 +4)))^2 ))∣=(1/2)a(√(a^2 +4))  α=tan^(−1) ((1/2)a(√(a^2 +4))) ■

Commented byamir last updated on 07/Feb/17

Commented byamir last updated on 07/Feb/17

dear mrW1! please draw a nice diagram  with geogebra and post it here.tnx.

Commented bymrW1 last updated on 08/Feb/17

Commented byamir last updated on 08/Feb/17

thank you very much mrw1.  you are a real mathmatic man.  goodlock!

Answered by mrW1 last updated on 03/Feb/17

to question 3:    if ∣a∣≤2, there are 2 orthogonal lines  from A. the area between them and  the curve is 0.    if ∣a∣>2, there are 4 orthogonal lines  from A.  F_1 =area of ARPB′  F_2 =area of ARQB  F_3 =area of B′BQP  F=area between orthogonal lines           and the curve, ABB′.  F=F_1 −F_2 −F_3   F_1 =(1/2)(a+y_4 )(a−x_4 )=(1/2)(a+(1/t_4 ))(a−t_4 )=(1/2)(a+(2/(a−(√(a^2 −4)))))(a+((a−(√(a^2 −4)))/2))  F_2 =(1/2)(a+y_3 )(a−x_3 )=(1/2)(a+(1/t_3 ))(a−t_3 )=(1/2)(a+(2/(a+(√(a^2 −4)))))(a+((a+(√(a^2 −4)))/2))  F_3 =∫_(x_4  ) ^x_3  (1/x)dx=(ln ∣x∣)∣_t_4  ^t_3  =ln ∣(t_3 /t_4 )∣=ln ∣((a+(√(a^2 −4)))/(a−(√(a^2 −4))))∣    F=(1/2)(a+(2/(a−(√(a^2 −4)))))(a+((a−(√(a^2 −4)))/2))−(1/2)(a+(2/(a+(√(a^2 −4)))))(a+((a+(√(a^2 −4)))/2))−ln ∣((a+(√(a^2 −4)))/(a−(√(a^2 −4))))∣