Consider a branch of the hyperbola $$x2$$−$$2y2$$−$$22x$$−$$42y$$−$$6=0$$ with vertex at the point A . Let B be one of the endpoints of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of triangle ABC is

Question

Consider a branch of the hyperbola $$x2$$−$$2y2$$−$$22x$$−$$42y$$−$$6=0$$ with vertex at the point A . Let B be one of  the endpoints of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of triangle ABC is

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$$x2$$−$$2y2$$−$$22x$$−$$42y$$−$$6=0$$ or  $$x2$$−$$22+2$$−$$2y2+22y+2=4$$ or  (x$$−$$2)24$$−$$(y+2)22=1Now$$ B is one of the end points of its latus rectum and C is the focus of the hyperbola nearest to the vertex A. Clearly, area of LABC does not change if we consider similar hyperbola with center at $$(0$$,$$0) or hyperbola $$x24$$−$$y22=1$$ Here vertex is $$A(2$$,$$0)$$ . ∴ $$a2e2=a2+b2=6$$ So, one of the foci is $$B(6$$,$$0)$$ , point C is ae,$$b2/a$$ or (6$$,$$1) . ∴ $$AB=6$$−$$2$$ and $$BC=1$$∴  Area of Δ$$ABC=12AB$$×$$BC=12(6$$−$$2)$$×$$1$$                                   $$=32$$−$$1$$ sq.units.

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

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Consider a branch of the hyperbola x2−2y2−22x−42y−6=0 with vertex at the point A . Let B be one of the endpoints of its latus rectum. If C is the focus of the hyperbola nearest to the point A, then the area of triangle ABC is

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