Hyperbola

General form of hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ ...(i) Given equation of hyperbola is $2x^{2} - y^{2} = 2$ $\Rightarrow x^{2} - \frac{y^{2}}{2} = 1$ ...(ii) comparing equation (i) with (ii), we get $a^{2} = 1, b^{2} = 2$ General equation of normal, $\frac{a^{2}x}{x_{1}} + \frac{b^{2}y}{y_{1}} = a^{2} + b^{2}$ ...(iii) Equation of normal at point $A(\sec \theta, 2 \tan \theta)$ $\frac{1 \times x}{\sec \theta} + \frac{2 \times y}{2 \tan \theta} = 1 + 2$ $\Rightarrow \frac{x}{\sec \theta} + \frac{y}{\tan \theta} = 3$ $\Rightarrow \frac{x}{\sec \theta} + \frac{y}{\frac{\sin \theta}{\cos \theta}} = 3$ $\Rightarrow \frac{x}{\sec \theta} + y \cos \theta \times \csc \theta = 3$ $\Rightarrow \frac{x}{\sec \theta} + \frac{y}{\sec \theta} \csc \theta = 3$ $\Rightarrow x + y \csc \theta = 3 \sec \theta$ ...(iv) Similarly, equation of normal at point $B(\sec \phi, 2 \tan \phi)$ $x + y \csc \phi = 3 \sec \phi$ $\Rightarrow x + y \csc\left(\frac{\pi}{2} - \theta\right) = 3 \sec\left(\frac{\pi}{2} - \theta\right)$ $\Rightarrow x + y \sec \theta = 3 \csc \theta$ ...(v) subtract equation (v) from (iv), we get $\Rightarrow y (\csc \theta - \sec \theta) = 3 (\sec \theta - \csc \theta)$ $\Rightarrow y (\csc \theta - \sec \theta) = -3 (\csc \theta - \sec \theta)$ $\Rightarrow y = -3$ Now, from equation (ii), we get $x^{2} - \frac{y^{2}}{2} = 1$ $x^{2} - \frac{9}{2} = 1$ $x^{2} = 1 + \frac{9}{2}$ $x = \sqrt{\frac{11}{2}}$ $\therefore (x, y) = (\alpha, \beta) = \left(\sqrt{\frac{11}{2}}, -3\right)$ $\therefore (2\beta)^{2} = (2 \times (-3))^{2} = 36$