Geo5.1.27

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Find the locus of the midpoints of chords of the parabola y^{2}=4ax\, which subtend a right angle at the vertex of the parabola.

Let P(x_{1},y_{1})\, be the midpoint of the chord BC of the parabola y^{2}=4ax\,

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

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

Join AB,AC. Given angle BAC is a right angle,homogenising y^{2}=4ax\,

y^{2}-4ax[{\frac  {yy_{1}-2ax}{y_{1}^{2}-2ax}}]=0\,

(y_{1}^{{2}}-2ax_{1})y^{2}-4ay_{1}xy+8a^{2}x^{2}=0\,

This is the combined equation of AB and AC. \angle BAC=90^{\circ }\,

coefficient of x^{2}\, + coefficient of y^{2}\,=0

8a^{2}+(y_{1}^{{2}}-2ax_{1})=0\,

y_{1}^{{2}}=2a(x_{1}-4a)\,

Therefore the locus of P is

y^{2}=2a(x-4a)\,


Main Page:Geometry:Plane|The Parabola