Geo5.1.31

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

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\,

Now

\tan \alpha =|{\frac  {2{\sqrt  {4a^{2}y_{1}^{{2}}-8a^{2}(y_{1}^{{2}}-2ax_{1})}}}{y_{1}^{{2}}-2ax_{1}+8a^{2}}}\,

Substitutingy_{1}^{{2}}=4ax_{1}\, in the denominator above and crossmultiplying and

squaring both sides,we have

\tan ^{{2}}\alpha (2ax_{1}+8a^{2})^{2}=|4(16a^{3}x_{1}-4a^{2}y_{1}^{{2}})|\,

\tan ^{{2}}\alpha (2ax_{1}+8a^{2})^{2}=16a^{2}(4ax_{1}-y_{1}^{{2}})\,

Simplifying further,

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

Therefore,the locus of P is

(x+4a)^{2}\tan ^{{2}}\alpha =4(y^{2}-4ax)\,


Main Page:Geometry:The Parabola