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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 end points of its latus rectum. If $C$ is the
focus of the hyperbola near $A,$ then area of the $∆ A B C$ is:

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32−1

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32+1.

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detailed solution

Correct option is B

Equation of the hyperbola is (x−2)24−(y+2)22=1Which can be written as X24−Y22=1 where x=X+2,y=Y−2. Eccentricity of the hyperbola is 4+24=32 Area of the ΔABC=12(AC)(BC)=12(ae−a)×b2a where a2=4,b2=2=12232−2×22=32−1

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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 end points of its latus rectum. If C is thefocus of the hyperbola near A, then area of the ∆ 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 end points of its latus rectum. If $C$ is the
focus of the hyperbola near $A,$ then area of the $∆ A B C$ is: