Consider a branch of the hyperbola $x^{2} - 2y^{2} - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$ with vertex at the point $A$. Let $B$ be one of the end points of its latus rectum. If $C$ is the focus of the hyperbola nearest to the point $A$, then the area of the triangle $ABC$ is

Consider a branch of the hyperbola x² - 2y² - 2√{2}x - 4√{2}y - 6 = 0 with vertex at the point A. Let B be one of the end points of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of the triangle ABC is $\text{Hyperbola: } x^{2} - 2y^{2} - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$ $A = \text{vertex},\quad B = \text{latus-rectum end},\quad C = \text{nearest focus to } A$ $\text{To find: area of } \triangle ABC \text{ from } A,\ B,\ C$ $\text{Options: } 1 \pm \sqrt{\tfrac{2}{3}},\quad \sqrt{\tfrac{3}{2}} \pm 1$ $x^{2} - 2\sqrt{2}x - 2\left(y^{2} + 2\sqrt{2}y\right) = 6$ $(x-\sqrt{2})^{2} - 2 - 2(y+\sqrt{2})^{2} + 4 = 6$ $(x-\sqrt{2})^{2} - 2(y+\sqrt{2})^{2} = 4$ $\dfrac{(x-\sqrt{2})^{2}}{4} - \dfrac{(y+\sqrt{2})^{2}}{2} = 1$ $\text{centre } (\sqrt{2}, -\sqrt{2}),\quad a^{2} = 4,\quad b^{2} = 2$ $c^{2} = a^{2} + b^{2} = 4 + 2 = 6 \Rightarrow c = \sqrt{6}$ $\text{foci: } (\sqrt{2} \pm \sqrt{6},\ -\sqrt{2})$ $A = (\sqrt{2}+2,\ -\sqrt{2}),\quad C = (\sqrt{2}+\sqrt{6},\ -\sqrt{2})$ $AC = c - a = \sqrt{6} - 2$ $\text{latus rectum} = \dfrac{2b^{2}}{a} = \dfrac{2 \cdot 2}{2} = 2$ $B = (\sqrt{2}+\sqrt{6},\ -\sqrt{2}+1)$ $BC = \dfrac{b^{2}}{a} = \dfrac{2}{2} = 1,\qquad BC \perp AC$ $\text{Area} = \dfrac{1}{2}\,(AC)(BC) = \dfrac{1}{2}(\sqrt{6}-2)(1)$ $= \dfrac{\sqrt{6}}{2} - 1 = \sqrt{\dfrac{3}{2}} - 1$ $\boxed{\text{Area} = \sqrt{\dfrac{3}{2}} - 1} \quad \Rightarrow \text{option (2)}$ $\text{Either branch gives the same area, by symmetry}$