A parabola has its vertex and focus in the first quadrant and axis along the line `y=x` . If the distances of the vertex and focus from the origin are respectively `sqrt(2)&2sqrt(2)` , then equation of the parabola is `x^2+y^2-8x+8y+2x y=16` `x^2+y^2-8x-8y+16=2x y` `(x-y)^2=8(x+y-2)` `(x+y)^2=(x-y+2)`

The axis of parabola is along the line y=x and the distance of its vertex and focus from 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:

The focus of the parabola x^2-8x+2y+7=0 is

The axis of a parabola is along the line y = x and its vertex and focus are in the first quadrant at distances sqrt2,2sqrt2 respectively, from the origin. The equation of the parabola, is

The equation of the directrix of the parabola x^(2) = 8y is

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)=4(x+y-2)(x-y)^(2)=4(x+y-2)(x-y)^(2)=4(x+y-2)(x-y)^(2)=8(x+y-2)