Consider a branch of the hyperbola x 2 - 2 y 2 - 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

Question

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 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 [2008]

Smart Achievers · Accepted Answer

(2)

$Given hyperbola is$

$x^{2} - 2 y^{2} - 2 \sqrt{2} x - 4 \sqrt{2} y - 6 = 0$

$\Rightarrow (x^{2} - 2 \sqrt{2} x + 2) - 2 (y^{2} + 2 \sqrt{2} y + 2) = 6 + 2 - 4$

$\Rightarrow {(x - \sqrt{2})}^{2} - 2 {(y + \sqrt{2})}^{2} = 4$

$\Rightarrow \frac{{(x - \sqrt{2})}^{2}}{2^{2}} - \frac{{(y + \sqrt{2})}^{2}}{(\sqrt{2})^{2}} = 1$

$∴$ $a = 2, b = \sqrt{2} \Rightarrow e = \sqrt{1 + \frac{2}{4}} = \sqrt{\frac{3}{2}}$

$Clearly ∆ A B C is a right triangle.$

$∴$ $Area (∆ A B C) = \frac{1}{2} \times A C \times B C$

$= \frac{1}{2} (a e - a) \times \frac{b^{2}}{a}$

$= \frac{1}{2} (e - 1) \times b^{2}$

$= \frac{1}{2} (\sqrt{\frac{3}{2}} - 1) \times 2$

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

Solution

Q.
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 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 [2008]

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

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

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

4 $\sqrt{\frac{3}{2}} + 1$

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Solution

Q. Consider a branch of the hyperbola x2-2y2-22x-42y-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 [2008]

1 1-23

2 32-1

3 1+23

4 32+1