VITEEE 2016 Paper

The equation of the hyperbola is $x^2 - 2y^2$ - 2x + 8y -1 = 0 or $(x-1)^2-2(y-2)^2$ + 6 = 0 or $\frac{(x-1)^2}{-6} + \frac{(y-2)^2}{3}$ = 1 or $\frac{(y-2)^2}{3} - \frac{(x-1)^2}{6}$ = 1 ... (1) or $\frac{Y^2}{3} - \frac{X^2}{6}$ = 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 = $\sqrt{3}$ , since in the equation (1) the transverse axis is parallel to the y-axis. If the conjugate axis is of length 2b, then b = $\sqrt{6}$ But $b^2$ = $a^2(e^2-1)$ ∴ 6 = $3(e^2-1)$ , ∴ $e^2$ = 3 or e = $\sqrt{3}$ The length of the transverse axis = $2\sqrt{3}$ The length of the conjugate axis = $2\sqrt{6}$ Latus rectum = $\frac{2b^2}{a}$ = $\frac{2 \times 6}{\sqrt{3}}$ = $4\sqrt{3}$