Find the extremities of latus rectum of the parabola

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Homework Help Overview

The problem involves finding the extremities of the latus rectum of the parabola defined by the equation y = (x^2) - 2x + 3. The original poster expresses difficulty in arriving at the correct answer and requests a full solution.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the transformation of the parabola into vertex form and the implications for identifying the latus rectum. Some participants question the original poster's assumptions about the coordinates of the latus rectum extremities and explore the relationship between the vertex and focus of the parabola.

Discussion Status

The discussion includes various attempts to clarify the correct form of the parabola and its properties. Some participants provide insights into the relationship between the vertex and focus, while others express uncertainty about the correct approach. There is no explicit consensus, but guidance has been offered regarding the definitions and properties of parabolas.

Contextual Notes

Participants note that the original poster is new to the topic and may be struggling with foundational concepts related to parabolas and their properties. There is an emphasis on understanding the geometric interpretations of the equations involved.

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


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..
 


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..
 


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...?
 


Hmmm. I don't know. Please you tell me
 


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?
 


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


Hey i got the correct ans. Thanks, thanks a lot for helping me
 
  • #10


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.
 

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