Geo5.1.30

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Show that the locus of the poles of chords which are normal to the parabola y^{2}=4ax\, is y^{2}(x+2a)+4a^{3}=0\,

Let P(x_{1},y_{1})\, be the pole of the normal chord

y=mx-2am-am^{3}\, w.r.t y^{2}=4ax\,

Equation of the polar of P(x_{1},y_{1})\, is

yy_{1}-2a(x+x_{1})=0\,

Both the first equation and this equation represent the same line

{\frac  {1}{y_{1}}}={\frac  {m}{2a}}={\frac  {-2am-am^{3}}{2ax_{1}}}\,

m={\frac  {2a}{y_{1}}},1={\frac  {-2a-am^{2}}{x_{1}}},x_{1}+2a+am^{2}=0\,

Eliminating m, we get

x_{1}+2a+a{\frac  {4a^{2}}{y_{1}^{{2}}}}=0\,

y_{1}^{{2}}(x_{1}+2a)+4a^{3}=0\,

Hence the locus of P is y^{2}(x+2a)+4a^{3}=0\,


Main Page:Geometry:The Parabola