Geo5.1.37

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Show that the locus of the poles of chords of the parabola y^{2}=4ax\, which are at constant distance 'd' from the focus is d^{2}(y^{2}+4a^{2})=4a^{2}(x+a)^{2}\,.

Let P(x_{1},y_{1})\, be the pole of a chord of the given parabola.

Equation to the polar of P with respect to the parabola is

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

Given perpendicular distance of the chord from the point S(a,0)\, is equal to d.

{\frac  {|-2a(x+x_{1})|}{{\sqrt  {y_{1}^{{2}}+4a^{2}}}}}=d\,

4a^{2}(x+x_{1})^{2}=d^{2}(y_{1}^{{2}}+4a^{2})\,

4(x^{2}+x_{1}^{{2}}+2axx_{1})=d^{2}(y_{1}^{{2}}+4a^{2})\,

Therefore,the locus of P is

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


Main Page:Geometry:The Parabola