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Prove that the locus of points the polars of which w.r.t the hyperbola {\frac  {x^{2}}{a^{2}}}-{\frac  {y^{2}}{b^{2}}}=1\, touch the circle x^{2}+y^{2}=c^{2}\, i f {\frac  {x^{2}}{a^{4}}}+{\frac  {y^{2}}{b^{4}}}={\frac  {1}{c^{2}}}\,

The polar of P w.r.t the hyperbola given is

{\frac  {xx_{1}}{a^{2}}}-{\frac  {yy_{1}}{b^{2}}}=1\,

This polar touches the circle x^{2}+y^{2}=c^{2}\, whose centre is (0,0) and radius c.


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

Squaring on both sides,

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

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

Therefore,the locus of the polars is

{\frac  {1}{c^{2}}}={\frac  {x^{2}}{a^{4}}}+{\frac  {y^{2}}{b^{4}}}\,

Main Page:Geometry:Hyperbola