Proving Locus of Middle Points of Chords on Parabola

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SUMMARY

The locus of the midpoints of all chords of the parabola defined by the equation y2=4ax that pass through the vertex is proven to be the parabola described by y2=2ax. The midpoint of a chord connecting the vertex and a point (h, 2√(ah)) on the parabola is calculated as (h/2, √(ah)). By eliminating the variable h, the relationship between x and y is established, confirming the locus as y2=2ax.

PREREQUISITES
  • Understanding of parabolic equations, specifically y2=4ax
  • Knowledge of coordinate geometry and midpoints
  • Ability to manipulate algebraic expressions and eliminate variables
  • Familiarity with the properties of parabolas
NEXT STEPS
  • Study the derivation of the locus of points for different conic sections
  • Learn about the properties of parabolas and their applications in geometry
  • Explore the concept of midpoints in coordinate geometry
  • Investigate the implications of locus in mathematical proofs
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Students studying coordinate geometry, mathematics educators, and anyone interested in the properties of parabolas and their applications in geometric proofs.

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Homework Statement


Prove that the locus of middle points of all chords of the parabola ##y^2=4ax## which are drawn through the vertex is the parabola ##y^2=2ax##.


Homework Equations





The Attempt at a Solution


The mid-point of the line joining the vertex and a point ##(h,2\sqrt{ah})## on parabola is ##(h/2,\sqrt{ah})## but what am I supposed to do with this? :confused:
 
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Pranav-Arora said:

Homework Statement


Prove that the locus of middle points of all chords of the parabola ##y^2=4ax## which are drawn through the vertex is the parabola ##y^2=2ax##.


Homework Equations





The Attempt at a Solution


The mid-point of the line joining the vertex and a point ##(h,2\sqrt{ah})## on parabola is ##(h/2,\sqrt{ah})## but what am I supposed to do with this? :confused:

You have ##x=h/2## and ##y=\sqrt{ah}##. Eliminate h?
 
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I like Serena said:
You have ##x=h/2## and ##y=\sqrt{ah}##. Eliminate h?

Woops, I was trying this question from the last night and I couldn't even notice that. Very foolish of me. :redface:

Thank you ILS! :smile:
 

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