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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..

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..

Grade:11

1 Answers

vikas askiitian expert
509 Points
10 years ago

extremites of lactus rectum are (2a,a)and(-2a,a) for parabola x^2=4ay

 extremitiesare (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)

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