Geo5.3.40

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P is any point on the hyperbola {\frac  {x^{2}}{a^{2}}}-{\frac  {y^{2}}{b^{2}}}=1\, whose vertex is A(a,0).Show that the locus of the middle point of AP is {\frac  {(2x-a)^{2}}{a^{2}}}-{\frac  {4y^{2}}{b^{2}}}=1\,

Let P(x_{1},y_{1})\, be the midpoint of AP of the given hyperbola

Equation of the chord AP is {\frac  {xx_{1}}{a^{2}}}-{\frac  {yy_{1}}{b^{2}}}={\frac  {x_{1}^{{2}}}{a^{2}}}-{\frac  {y_{1}^{{2}}}{b^{2}}}\,

A(a,0)\, lies on the above equation,hence

{\frac  {ax_{1}}{a^{2}}}-0={\frac  {x_{1}^{{2}}}{a^{2}}}-{\frac  {y_{1}^{{2}}}{b^{2}}}\,

{\frac  {x_{1}^{{2}}}{a^{2}}}-{\frac  {x_{1}}{a}}-{\frac  {y_{1}^{{2}}}{b^{2}}}=0\,

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

{\frac  {4x_{1}^{{2}}-4ax_{1}+a^{2}}{a^{2}}}-{\frac  {4y^{2}}{b^{2}}}=1\,

{\frac  {(2x_{1}-a)^{2}}{a^{2}}}-{\frac  {4y_{1}^{{2}}}{b^{2}}}=1\,

Therefore,locus of P is {\frac  {(2x-a)^{2}}{a^{2}}}-{\frac  {4y^{2}}{b^{2}}}=1\,


Main Page:Geometry:Hyperbola