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  • Consider a branch of the hyperbola <img alt="\mathrm {x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%20%7Bx%5E2-2%20y%5E2-2%20%5Csqr

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

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

The hyperbola \mathrm {x^2-2 y^2-2 \sqrt{2} x-4 \sqrt{2} y-6=0}  can be reduced to 

\mathrm {\frac{(x-\sqrt{2})^2}{4}-\frac{(y+\sqrt{2})^2}{2}=1 \: \: We \: \: have \: \: a=2 \quad b=\sqrt{2}}


\mathrm{ e=\sqrt{\frac{b^2}{a^2}+1}=\sqrt{\frac{3}{2}} }


The area of the triangle \mathrm { A B C }


\mathrm { =\frac{1}{2} a(e-1) \cdot \frac{b^2}{a}=\frac{b^2(e-1)}{2}=\frac{2\left(\sqrt{\frac{3}{2}}-1\right)}{2}=\sqrt{\frac{3}{2}}-1 }

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seema garhwal

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