Geo5.1.29

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Prove that the locus of the midpoints of the focal chords of the parabola y^{2}=4ax\, is another parabola whose vertex is the focus of y^{2}=4ax\,.

Given the parabola is y^{2}=4ax\,. Let (x_{1},y_{1})\, be the midpoint of a focal chord.

Equation of the chord with (x_{1},y_{1})\, as midpoint is

yy_{1}-2a(x+x_{1})=y_{1}^{{2}}-4ax_{1}\,

yy_{1}-2ax=y_{1}^{{2}}-2ax_{1}\,

Since this equation passes thtough the focus (a,0)\,, we have

y_{1}^{{2}}=2a(x_{1}-a)\,

Therefore, the locus of the midpoint is y^{2}=2a(x-a)\,

This is a parabola with vertex (a,0)\, which is the focus of the given parabola.


Main Page:Geometry:The Parabola