Factorise the expression: `(x^2–2xy+y^2)–z^2`

Answer

B

Explanation

`x^(2)-2y^(2)-2sqrt2x-4sqrt2y-6=0`
`"or "(x^(2)-2sqrt2x+3)-2(y^(2)+2sqrt2y+2)=4`
`"or "((x-sqrt2)^(2))/(4)-((y-sqrt2)^(2))/(2)=1`
Now, 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 `Delta`ABC does not change if we consider similar hyperbola with centre at (0, 0) or hyperbola `(x^(2))/(4)-(y^(2))/(2)=1`

Here vertex is A(2, 0).
`therefore" "a^(2)e^(2)=a^(2)+b^(2)=6`
So, one of the foci is `B(sqrt6,0)`. point C is `(ae, b^(2)//a) or (sqrt6,1)`.
`therefore" "AC=sqrt6-2 and BC=1`
`therefore" Area of "DeltaABC=(1)/(2)xxACxxBC=(1)/(2)(sqrt6-2)xx1`
`=sqrt((3)/(2))-1" sq. units."`