The equation of a parabola is `(x + y - 2)^2 = sqrt2 (x - y)` then Vertex of the parabola is `(1,1)`

The equation of a parabola is (x + y - 2)^2 = sqrt2 (x - y) then (a) Vertex of the parabola is (1,1) (b) latus rectum of parabola is 2. (c) The equation of tangent at vertex is x+y-2=0. (d) the perpendicular from focus on any tangent of the parabola is x-y=0.

Find the vertex of the parabola x^2=2(2x+y) .

Find the vertex of the parabola x^2=2(2x+y) .

The vertex of the parabola x^(2)=8y-1 is :

The vertex of the parabola y^2=4a(x+a) is:

Find the vertex of the parabola x^2=2(2x+y)

The vertex of the parabola y^(2)=4 a(x+a) is

Find the vertex of the parabola x^(2)=2(2x+y)

The vertex of a parabola is (a,0) and the directrix is x+y=3a the equation of the parabola is