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Mathematics Dictionary
Dr. K. G. Shih
Figure 318 : R = (D*e)/(1 + e*cos(A))

  • Q01 | - Diagram : R = (D*e)/(1 + e*cos(A))
  • Q02 | - Compare R = (D*e)/(1 + e*cos(A)) with R = (D*e)/(1 + e*cos(A))
  • Q03 | - Compare R = (D*e)/(1 + e*cos(A)) with (x/a)^2 - (y/b)^2 = 1
  • Q04 | - Compare x^2 - y^2 = 1 with R = 1/(1 - Sqr(2)*cos(A))
  • Q05 | - Hyperbola x*y = 1
  • Q06 | - Equation of directrix of Hyperbola x*y = 1

Q01. Diagram :


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Q02. Compare R = (D*e)/(1 + e*cos(A)) with R = (D*e)/(1 - e*cos(A))

R = (D*e)/(1 - e*cos(A))
  • Origin at focus G(0,0)
  • Center is at (-f, 0)
  • Directrix is at left of focus G
  • ((x + f)/x)^2 + (y/b)^2 = 1
  • Angle A = 180, R= GV = (D*e)/(1 + e)
R = (D*e)/(1 + e*cos(A))
  • Origin at focus F(0,0)
  • Center is at (f, 0)
  • Directrix is at right of focus F
  • ((x - f)/x)^2 + (y/b)^2 = 1
  • Angle A = 0, R = (D*e)/(1 + e)

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Q03. Compare R = (D*e)/(1 + e*cos(A)) with (x/a)^2 - (y/b)^2 = 1

(x/a)^2 - (y/b)^2 = 1
  • Center at (0, 0)
  • f = Sqr(a^2 + b^2)
  • Vertices : (-a, 0) and (a, 0)
  • Foci : (-f, 0) and (f, 0)
  • e = f/a
  • D*e = (b^2)/a
R = (D*e)/(1 + e*cos(A))
  • Origin at focus F(0, 0)
  • Angle A = 0
    • cos(A) = 1
    • FU = f - a
    • R = FU = (D*e)/(1 + e)
    • D*e = FU*(1 + e)
    • D*e = (f - a)*(f + a)/a
    • D*e = (f^2 - a^2)/a
    • Since b^2 = f^2 - a^2
    • Hence D*e = (b^2)/a
  • Vertex U at (a, 0)
  • Center C at (f, 0)
Conculsion
  • R = (D*e)/(1 + e*cos(A)) is same as (x/a)^2 + (y/b)^2 = 1
  • The only differencs are the coordinates of vertex and center
  • But we can translater (x/a)^2 + (y/b)^2 = 1 to match the coordinates

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Q04. Compare x^2 - y^2 = 1 with R = 1/(1 - Sqr(2)*cos(A))

Hyperbola : x^2 - y^2 = 1
  • a = 1 and b = 1
  • f = Sqr(a^2 + b^2) = Sqr(2)
  • e = f/a = Sqr(2)
  • D*e = (b^2)/a = 1
  • Since corresponding polar form is R = (D*e)/(1 - e*cos(A))
  • Hence R = 1/(1 - Sqr(2)*cos(A)) is congruent to x^2 - y^2 = 1

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Q05. Hyperbola x*y = 1

Question
  • 1. Find semi-axese
  • 2. Find focal length
  • 3. Find corrdinates of foci
Rotation matrix
  • | +cos(A) +sin(A) |
  • | -sin(A) +cos(A) |
Rotation coordinate x and y to coordiante to u and v
  • u = +x*cos(A) + y*sin(A)
  • v = -x*sin(A) + y*cos(A)
Rotation coordinate u and v to coordiante to x and y
  • x = +u*cos(A) - v*sin(A)
  • y = +u*sin(A) + v*cos(A)
Change x*y = 1 to u and v coordiante by rotating 45 degrees
  • x*y = (+u*cos(A) - v*sin(A))*(+u*sin(A) + v*cos(A)) = 1
  • Since cos(45) = Sqr(2)/2 and sin(45) = Sqr(2)/2
  • Hence (Sqr(2)*u/2 - Sqr(2)*v/2)*(Sqr(2)*u/2 + Sqr(2)*v/2) = 1
  • Since (x - y)*(x + y) = x^2 - y^2
  • Hence (1/2)*(u^2 - v^2) = 1
  • Hence (u/Sqr(2))^2 - (v/Sqr(2))^2 = 1
  • Hence a = b = Sqr(2)
  • Hence f = Sqr(a^2 + b^2) = 2
Find coordinates of foci of x*y = 1
  • Vertices of u^2 - v^2 = 2
    • (-Sqr(2), -Sqr(2)) and (Sqr(2), Sqr(2))
  • Vertices of x*y = 1
    • (-1, 1) and (1, 1)
  • Foci of u^2 - v^2 = 2
    • (-2, -2) and (2, 2)
  • Foci of x*y = 1
    • (-Sqr(2), -Sqr(2)) and (Sqr(2), Sqr(2))

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Q06. Equation of directrix of Hyperbola x*y = 1

Question
  • 1. Compare x*y = 1 with (x/Sqr(2))^2 - (y/Sqr(2))^2 = 1
  • 2. Find equation of directrix
Solution using diagram
  • (x/Sqr(2))^2 - (y/Sqr(2))^2 = 1
    • Principal axis y = 0
    • a = Sqr(2) and b = Sqr(2)
    • Center at (0, 0)
    • Vertex U at (-Sqr(2), 0)
    • Focal length f = Sqr(a^2 + b^2) = 2
    • D*e = (b^2)/a = 1
    • e = f/a = Sqr(2)
    • Hence D = Sqr(2)/2 = distance from focus F to directrix
  • Diagram x*y = 1
    • Vertex U(1, 1)
    • Center C(0, 0)
    • Principal axis is y = x
    • CU = Sqr(xu^2 + yu^2) = Sqr(2)
    • Hence a = Sqr(2) and b = Sqr(2)
    • Hence CF = f = Sqr(a^2 + b^2) = 2
    • Hence Focus F at (-Sqr(2), -Sqr(2))
    • D*e = (b^2)/a = 1 and e = f/a = Sqr(2)
    • Hence D = Sqr(2)/2
  • Conclusion
    • x*y = 1 is same as (x/Sqr(2))^2 - (y/Sqr(2))^2 = 1
    • The difference is the principal axese
      • For (x/a)^2 - (y/b)^2 = 1 is y = 0
      • For x*y = 1 is y = -x
Find equation of directix x*y = 1
  • Equation directrix is y = -x + c
  • CF = CQ + QF and QF = D is the distance from F to directrix
  • CQ = CF - QF = 2 - Sqr(2)/2 = 1.2929
  • Find c using triangle right triangle CQP
    • Let P be intersection of directrix with y-axis
    • Since angle CQP = 90 and angle CPQ = 45 degrees
    • Hence QP = CQ = 1.2929
    • Hence CP = Sqr(PQ^2 + CQ^2)
    • = Sqr(2*(1.2929)^2)
    • = Sqr(3.34318)
    • = 1.828436
    • Hence c = -1.828436
    • Hence y = - x - 1.828436
  • 2nd method to find c : Using coordinate of Q (-0.9142, -0.9142)
    • Since CQ = 1.2929
    • Hence xq^2 + yq^2 = CQ^2
    • Since xq = yq
    • Hence xq = -0.9142 and yq = -0.9142
    • Substitute xq, yq into y = -x + c
    • Hence c = -18284

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