The equation of the hyperbola is
x2−2y2 - 2x + 8y -1 = 0
or
(x−1)2−2(y−2)2 + 6 = 0
or
+ = 1
or
− = 1 ... (1)
or
− = 1
where X = x-1 and Y = y-2 ...(2)
∴ The centre = (0,0) in the X-Y co-ordinates.
∴ The centre = (1,2) in the x-y co-ordinates, using (2).
If the transverse axis be of length 2a, then a =
√3 , since in the equation (1) the transverse axis is parallel to the y-axis. If the conjugate axis is of length 2b, then
b =
√6 But
b2 =
a2(e2−1) ∴ 6 =
3(e2−1) , ∴
e2 = 3 or e =
√3 The length of the transverse axis =
2√3 The length of the conjugate axis =
2√6 Latus rectum =
=
=
4√3