coor from jee 2006

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 _____

5 Answers

1357
Manish Shankar ·

one 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)

1
Akand ·

distance 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????

1
Akand ·

shud 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

3
msp ·

than q sir .and akand i have done in that way only day and my luk says only one option satisfid .

1
Akand ·

hehe nice..........it always works..[3][3][3]

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