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Consider a branch of the hypebola x^2-2y...

Consider a branch of the hypebola `x^2-2y^2-2sqrt2x-4sqrt2y-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

A

`1-sqrt((2)/(3))`

B

`sqrt((3)/(2)-1`

C

`1+sqrt((2)/(3))`

D

`sqrt((3)/(2)+1`

Text Solution

Verified by Experts

The correct Answer is:
B
The equation of the hyperbola is
`(x-sqrt(2))^(2)-2(y+sqrt(2))^(2)=1`
`implies((x-sqrt(2))^(2))/(2^(2))(sqrt(2))^(2)(y+sqrt(2))^(2)=1`
Its centre is at `(sqrt(2),-sqrt(2))` and eccentricity `e` is given by
`e=sqrt(1+(2)/(4)=sqrt((3)/(4))`
The coordinates of the vertex `A` are given by
`x-sqrt(2)=2` and `y+sqrt(2)=0`
`impliesx=2+sqrt(2)`, `y=-sqrt(2)`
So, coordinates of vertex `A` are `(2+sqrt(2),-sqrt(2))`
The coordinate of the focus `C` nearest to vertex `A` are given by
`x-sqrt(2)=2xxsqrt((3)/(2))`, `y+sqrt(2)=0impliesx=sqrt(2)+sqrt(6)`, `y=-sqrt(2)`
Thus, the coordinates of focus `C` are `(sqrt(2)+sqrt(6),-sqrt(2))`
Length of the latusrectum `=2xx(2)/(2)=2`
`:.` Coordinate of `B` are `(sqrt(2)+sqrt(6),1)`.
Clearly, `DeltaABC` is a right triangle right angled at `B`.
`:.` Area of `DeltaABC=(1)/(2)(ABxxBC)`
`implies` Area of `DeltaABC=(1)/(2)|(sqrt(2)+sqrt(6))-(2+sqrt(2))|xx1`
`implies` Area of ` DeltaABC=(sqrt(6)-2)/(2)=(sqrt(3))/(2)-1`
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