Geo5.1.34

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Show that the locus of the midpoints of chords of the prabolay^{2}=4ax\, which touch the circle x^{2}+y^{2}=a^{2}\, is (y^{2}-2ax)^{2}=a^{2}(y^{2}+4a^{2})\,

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

Equation of the chord is

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

-2ax+yy_{1}+2ax_{1}-y_{1}^{{2}}=0\,

2ax-yy_{1}+y_{1}^{{2}}-2ax_{1}=0\,

This line touch the circle with centre (0,0) and radius a.

The perpendicualar distance from the centre is equal to the readius,

Therefore,

{\frac  {|y_{1}^{{2}}-2ax_{1}|}{{\sqrt  {y_{1}^{{2}}+4a^{2}}}}}=a\,

Squaring on bothsides,we get

[y_{1}^{{2}}-2ax_{1}]^{2}=a^{2}(y_{1}^{{2}}+4a^{2})\,

Hence the locus of P is

[y^{2}-2ax]^{2}=a^{2}(y^{2}+4a^{2})\,


Main Page:Geometry:The Parabola