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The axis of a parabola is along the line...

The axis of a parabola is along the line `y=x` and the distance of its vertex and focus from the origin are `sqrt(2)` and `2sqrt(2)` , respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is `(x+y)^2=(x-y-2)` `(x-y)^2=(x+y-2)` `(x-y)^2=4(x+y-2)` `(x-y)^2=8(x+y-2)`

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The correct Answer is:
A
Since, distance of vertex from origin is `sqrt(2)` and focus is `2 sqrt(2)`
`:. ` V (1,1) and f (2,2) (i.e. lying on y = x)
where, length of latusretum
`= 4a = 4 sqrt(2)`
`:.` By definition of parabola

`PM^(2) = (4a) PN)`
Where, PN is length of perpendicular upon
`x + y - 2 = 0`, i.e., tangent at vertex
`implies ((x - y)^(2))/(2) = 4 sqrt(2) ((x + y - 2)/(sqrt(2)))`
`implies (x - y)^(2) = 8 (x + y - 2)`
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