Geo5.1.32

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Show that the locus of the poles of the chords of the parabola y^{2}=4ax\, which subtend a right angle at the vertex is (x+4a)=0\,

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=\alpha \,

Homogenising y^{2}=4ax\,

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

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

Since the angle at the vertex is right angle,

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

Hence

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

Substituting y_{1}^{{2}}=4ax_{1}\,

8a^{2}+2ax_{1}=0\,

4a+x_{1}=0\,

Hence the locus of P is

x+4a=0\,

Main Page:Geometry:The Parabola