VITEEE 2016 Paper

The equation of the hyperbola is
x2−2y2 - 2x + 8y -1 = 0
or (x−1)2−2(y−2)2 + 6 = 0
or

(x−1)2

−6

+

(y−2)2

3

= 1
or

(y−2)2

3

−

(x−1)2

6

= 1 ... (1)
or

Y2

3

−

X2

6

= 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 =

2b2

a

=

2×6

√3

= 4√3