Parabola

[rajatgl16]

Find the extremities of latus rectum of the parabola y=(x^2)-2x+3.

Please someone post its solution. Ans. is (1/2,9/4) (3/2,9/4).

i just need full solution. I tried a lot but didn't get this correct answer.

[Mentallic]

Show us what you've tried in order to find the equation of the latus rectum.

[rajatgl16]

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..

[rajatgl16]

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..

[Mentallic]

They should be those coordinates for a parabola that is y=x2 but yours isn't that, it is y-2=(x-1)2. for a parabola y=x2 the vertex is at (0,0) and focus is at (0,a) which suggests that for a parabola is the form y-k=(x-h)2 the vertex is at (h,k) and the focus is then at...?

[rajatgl16]

Hmmm. I dont know. Please you tell me

[Mentallic]

If the vertex is at (0,0) and the focus is at (0,a) then the focus is always a units above the vertex (actually, inside the parabola would be better since if the parabola is curving downwards then the focus is a units down). Then for a parabola with centre (h,k) the focus will be?

[rajatgl16]

Focus should be (h, k+a). And what about latus rectum

[rajatgl16]

Hey i got the correct ans. Thanks, thanks a lot for helping me

[Mentallic]

That's it! :smile: The latus rectum is just y=k+a, so you find where that line intercepts the parabola.

No problem, take care.