Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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 parabola
A)(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)
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
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !