Geo5.1.23

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Show that the equation of the chord joining the points (x_{1},y_{1}),(x_{2},y_{2})\, on the parabola y^{2}=4ax\, is (y-y_{1})(y-y_{2})=y^{2}-4ax\,

Given the parabola S\equiv y^{2}-4ax=0\,

Equation to the chord joining (x_{1},y_{1}),(x_{2},y_{2})\, on the parabola is

S_{1}+S_{2}=S_{{12}}\,

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

yy_{1}+yy_{2}-4ax=y_{1}y_{2}\,

-4ax=-y(y_{1}+y_{2})+y_{1}y_{2}\,

Adding y^{2}\, on bothsides,

y^{2}-4ax=y^{2}-y(y_{1}+y_{2})+y_{1}y_{2}\,

y^{2}-4ax=(y-y_{1})(y-y_{2})\,

Main Page:Geometry:The Parabola