Find the extremities of latus rectum of the parabola y=(x^2)-2x+3. Ans. is (1/2,9/4) (3/2,9/4). i just need full solution. i tried it as follo:

y=x^2-2x+3
x^2-2x=y-3
x^2-2x+1=y-3+1
(x-1)^2=1(y-2)

let X=x-1 and Y=y-2
then equation becomes X^2=Y

on comparing it with x^2=4ay which is general form of parabolic equations we get
a=1/4

so according to me extremities of latus rectum should be (0,1/4) (0,-1/4)....


I'm in 11th grade and i just started this topic first time few days back. So i'm very new for it..

Rajat Goel

15 Years agoGrade 11