Geo5.1.28

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Show that the locus of the midpoints of chords of y^{2}=4ax\, which subtend a constant angle alpha at the vertex is [y^{2}-2ax+8a^{2}]^{2}=16a^{2}\cot ^{{2}}\alpha (4ax-y^{2})\,.

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\, with the equation of BC, we get

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\,

Hence the angle between the lines is

\cot \alpha =|{\frac  {y_{1}^{{2}}-2ax_{1}+8a^{2}}{{\sqrt  {(y_{1}^{{2}}-2ax_{1}-8a^{2})^{2}+16a^{2}y_{1}^{{2}}}}}}|\,

Simplifying the equation we get

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

Hence the locus of P is

16a^{2}\cot ^{{2}}\alpha (4ax-y^{2})=(y^{2}-2ax+8a^{2})^{2}\,


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