θ θ 45 - θ θ b a r s Mirror diameter = D Focal ratio = F Focal length = a + b Major axis = r + s tan θ = D / 2(a+b) = 1 / 2F D b tan θ

Problem 1. Determine the semi-major axis of the secondary ellipse. Solution: Law of sines: sin(45-θ) / b = sin θ / r sin(135-θ) / b = sin θ / s r + s = b sin θ ( 1 / sin(45 – θ) + 1 / sin(135 – θ) ) = sqrt(2) b ( 2 sin θ cos θ / (cos^2 θ – sin^2 θ) ) = 2 sqrt(2) b ( tan θ / ( 1 – tan^2 θ ) ) = 2 sqrt(2) b ( 2F / (2F + 1)(2F – 1) ) So Semi-major axis = ½ (r+s) = sqrt(2) b * 2F / (2F + 1)(2F – 1)

b a ½ (r+s) D Offset Drop m s Definition of offset and drop: r = ½ (r+s) + m or m = ½ (r-s) Drop = offset = m / sqrt(2)

Problem 2. Determine the offset and drop. Solution: Law of sines again: sin(45-θ) / b = sin θ / r sin(135-θ) / b = sin θ / s r - s = b sin θ ( 1 / sin(45 – θ) - 1 / sin(135 – θ) ) (r – s) / 2 sqrt(2) = b sin^2 θ / ( cos^2 θ - sin^2 θ ) = b / ( 1 / tan^2 θ - 1 ) = b / (4F^2 – 1) So Offset = b / (4F^2 – 1) = b / (2F-1)(2F+1) In order that the offset be physically positive, the focal ratio must be greater than ½. Thus f = ½ represents a singular case.

Problem 3. Determine the minor axis and eccentricity of the diagonal ellipse. Solution: Refer to the diagram which represents the secondary diagonal seen face on. The semi-minor axis is B. We take b tan θ from the original diagram, because the cone has a circular base. s ½ (r+s) B b tan θ m The length m is sqrt(2) * offset = sqrt(2) b / (2F+1)(2F-1)

All points on the diagonal ellipse satisfy the equation: (x/A)^2 + (y/B)^1 = 1 Since we know one such point as shown, and A, we can find B. b tan θ m B A x y

A = ½ (r+s) = sqrt(2) b * 2F / (2F + 1)(2F – 1) x = -m = - sqrt(2) b / (2F+1)(2F-1) y = b tan θ = b / 2F (x/A)^2 = (-1/2F)^2 = 1/4F^2 (y/B) = sqrt(1 – (x/A)^2) = sqrt(1 – 1/4F^2) B = y / sqrt(1 – 1/4F^2) = b / sqrt(4F^2 – 1) = b / sqrt( (2F+1)(2F-1) ) Semi-minor axis = b / sqrt( (2F+1)(2F-1) )

Problem 4. Determine the eccentricity of the diagonal ellipse. Solution: The eccentricity is defined to be e = sqrt ( 1 – (B/A)^2 ) = 1/sqrt(2) * sqrt( 1 + (1/2F)^2 ) = 1/sqrt(2) * sec θ where θ is ½ the angle of the mirror subtends at its focal point. At the singular value F = ½ the eccentricity is 1 and the ellipse goes over into a parabola.