Find the extremities of Latus Rectum of the parabola y= x2 - 2x + 3 .
y=x2-2x+3
x2-2x=y-3
x2-2x+1=y-3+1
(x-1)2=1(y-2)
let X=x-1 and Y=y-2
then equation becomes X2=Y
on comparing it with x2=4ay which is general form of parabolic equations we get
a=1/4.
extremites of lactus rectum are (2a,a)and(-2a,a) for parabola x2=4ay
extremities are (1/2,1/4) and (-1/2,1/4) as a=1/4
but these points are with respect to coordinate system having (x,y) as general point
therefore (x-1,y-2) = (X,Y) =(1/2,1/4),(-1/2,1/4)
from here we get (x,y)=(1/2,9/4),(3/2,9/4)
x2-2x=y-3
x2-2x+1=y-3+1
(x-1)2=1(y-2)
let X=x-1 and Y=y-2
then equation becomes X2=Y
on comparing it with x2=4ay which is general form of parabolic equations we get
a=1/4.
extremites of lactus rectum are (2a,a)and(-2a,a) for parabola x2=4ay
extremities are (1/2,1/4) and (-1/2,1/4) as a=1/4
but these points are with respect to coordinate system having (x,y) as general point
therefore (x-1,y-2) = (X,Y) =(1/2,1/4),(-1/2,1/4)
from here we get (x,y)=(1/2,9/4),(3/2,9/4)