Geo5.1.38

From Example Problems
Jump to: navigation, search

Show that the locus of the midpoints of chords of the parabola y^{2}=6x\, and which touch the circle x^{2}+y^{2}+4x-12=0\, is [y^{2}-3x-6]^{2}=16(y^{2}+9)\,

Equation of the midpoints of chords of the given parabola if P(x_{1},y_{1})\, is the midpoint,

yy_{1}-3(x+x_{1})=y_{1}^{{2}}-6x_{1}\,

yy_{1}-3x=3x_{1}+y_{1}^{{2}}-6x_{1}\,

This line touches the circle with centre (-2,0)\, and radius is 4.

Hence the distance from the centre to the line is

{\frac  {|6-y_{1}^{{2}}+3x_{1}|}{{\sqrt  {y_{1}^{{2}}+9}}}}=4\,

Squaring on bothsides,we get

[y_{1}^{{2}}-3x_{1}-6]^{{2}}=16(y_{1}^{{2}}+9)\,

Therefore,the locus is

[y^{2}-3x-6]^{{2}}=16(y^{2}+9)\,

Main Page:Geometry:The Parabola