If distance of vertex and foci from origin are `'a' and 'a^1'` on +x axis of parabola will be `y^2=4(a^1-a)(x-a)` such that `a^1>a.`

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

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 the equation of the parabola is :

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 the equation of the parabola is :

The axis of a 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 find equation of the parabola.

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:

Prove that the equation of the parabola whose vertex and focus are at distances alpha and beta from origin on x-axis respectively is y^2=4(beta-alpha)(x-alpha) .

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)

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

The axis of a parabola is along the line y=x and the distance of its vertex from origin is sqrt(2) and that from its focus is 2sqrt(2) if vertex and focus both lie in the first quadrant then the equatio of the parbola is