The axis of parabola is along the line y=x and the distance of its vertex and focus from origin are `sqrt2` and 2`sqrt2` respectively. If vertex and focus both lie in the first quadrant, then find the equation of the parabola

Answer

`(x-y)^(2)=8(x+y-2)`

Explanation

Axis of parabola is y=x.

Sinec vertex is at distance `sqrt(2)` from (0,0), vertex is A(1,1).
Also, focus is at distance `2sqrt(2)` from (0,0). So, focus is S(2,2).
Distance `SA=sqrt(2)`
So, directrix is at distance `sqrt(2)` from origin, which is x+y=0.
Hence, equation of the parabola is
`sqrt((x-2)^(2)+(y-2)^(2))=(|x+y|)/(sqrt(2))`
`orx^(2)+y^(2)-2xy=8(x+y-2)`
`or(x-y)^(2)=8(x+y-2)`