The axis of a parabola is along the line y=x,and the distance of its vertex from origin is √2 and that from its frocus is 2√2 both lie in the first quadrant,then the equation of the parabola is _____
1357one of the way is
vertex is (1,1) focus is (2,2)
directix is y=-x that is x+y=0 (same distance from vertex as focus and perpendicular to axis)
so equation is
distance of point from focus=distance of the point from directrix
(h-2)2+(k-2)2=(h+k)2/2
solving it we get
(h-k)2=8(h+k-2)
(x-y)2=8(x+y-2)
1distance of its vertex from origin is √2 and that from its focus is 2√2
is d distance from d vertex to d focus 2√2 or is it d distance from d focus to d origin????
1shud i give an easier and a quicker way?????
wel after finding d vertex as (1,1) in d first step......substitute it in d options , if ur luck is gud 3 will not satisfy .....eliminate all dose.....
if u hav a confusion between 2 options find its focus....if tht comes out to be (2,2)....then its solved.....
THIS is a tip for not wasting time by doing d question in d standard way heheh
3than q sir .and akand i have done in that way only day and my luk says only one option satisfid .