 # axis of a parabola is y=x and vertex and focus are at distance 2^(1/2) and 2*2^(1/2) respectivaly from the origin. then equation pf parabolaA)(x-y)^2=8*(x+y-2)B) (x+y)^2=2*(x+y-2)c) (x-y)^2=4*(x+y-2)D) (x+y)^2=2*(x-y+2)

12 years ago

distance of vertex from focus & directrix is same  ...if parabola is considered in +xy plane then

vertex is at 21/2 away from origin ....

(x2,y2) are coordinates of vertex then

y2-y1/sin@ = x2-x1/cos@ = d             ................1

d = 21/2 , (x1,y1) = (0,0)  & cos@ = 1/21/2  (coz slope is 1) so

y2 = 1 , x2 = 1

vertex = (1,1)

now focus by same formula 1 will be

focus = (2,2)

directrix will be perpendicular line to axis of parabola passing through origin

y = -x will be directrix

distance of focus from directrix = distance of any point on parabola to focus

(y+x)/root2 = [ (x-2)2+(y-2)2 ]1/2

y2 + x2 + 2xy = 2 [ x2 + y2 - 4x - 4y + 8 ]

x2+y2 -2xy =  (8x+8y-16)

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

this is the required eq ... option A) is correct

approve if u like my ans