1.Axis of a parabola is y=xy = xy=x and vertex and focus are at a distance 2\sqrt{2}2 and 222\sqrt{2}22 respectively, from the origin. Then, equation of the parabola isa.(x−y)2=8(x+y−2)(x - y)^2 = 8(x + y - 2)(x−y)2=8(x+y−2)b.(x+y)2=2(x+y−2)(x + y)^2 = 2(x + y - 2)(x+y)2=2(x+y−2)c.(x−y)2=4(x+y−2)(x - y)^2 = 4(x + y - 2)(x−y)2=4(x+y−2)d.(x+y)2=2(x−y+2)(x + y)^2 = 2(x - y + 2)(x+y)2=2(x−y+2)Login to continueOnly logged in users canattempt or see the solution.
