Consider a branch of the hyperbola x^2 - 2y^2 | vedclass

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(B) The given equation is $x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$.Completing the square,we get $(x^2 - 2\sqrt{2}x + 2) - 2(y^2 + 2\sqrt{2}y + 2

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$\sqrt{\frac{3}{2}} - 1$

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(B) The given equation is $x^2 - 2y^2 - 2\sqrt{2}x - 4\sqrt{2}y - 6 = 0$.Completing the square,we get $(x^2 - 2\sqrt{2}x + 2) - 2(y^2 + 2\sqrt{2}y + 2) = 6 + 2 - 4$,which simplifies to $(x - \sqrt{2})^2 - 2(y + \sqrt{2})^2 = 4$.Dividing by $4$,we obtain the standard form: $\frac{(x - \sqrt{2})^2}{4} - \frac{(y + \sqrt{2})^2}{2} = 1$.Here,$a^2 = 4 \Rightarrow a = 2$ and $b^2 = 2 \Rightarrow b = \sqrt{2}$.The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{2}{4}} = \sqrt{\frac{3}{2}}$.The vertex $A$ is $(\sqrt{2} + a, -\sqrt{2}) = (2 + \sqrt{2}, -\sqrt{2})$.The focus $C$ nearest to $A$ is $(\sqrt{2} + ae, -\sqrt{2}) = (\sqrt{2} + 2\sqrt{\frac{3}{2}}, -\sqrt{2}) = (\sqrt{2} + \sqrt{6}, -\sqrt{2})$.The distance $AC = ae - a = a(e - 1) = 2(\sqrt{\frac{3}{2}} - 1) = \sqrt{6} - 2$.The length of the semi-latus rectum is $\frac{b^2}{a} = \frac{2}{2} = 1$.The point $B$ is $(ae + \sqrt{2}, \frac{b^2}{a} - \sqrt{2}) = (\sqrt{6} + \sqrt{2}, 1 - \sqrt{2})$.The area of the triangle $ABC$ with base $AC$ and height $BC$ is $\frac{1}{2} 	imes AC 	imes \frac{b^2}{a} = \frac{1}{2} 	imes (\sqrt{6} - 2) 	imes 1 = \frac{\sqrt{6}}{2} - 1 = \sqrt{\frac{6}{4}} - 1 = \sqrt{\frac{3}{2}} - 1$.

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

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