Geo5.3.45

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If e1,e2 are the eccentricities of a hyperbola and its conjugate,prove that {\frac  {1}{e_{1}^{{2}}}}+{\frac  {1}{e_{2}^{{2}}}}=1\,


Let e1 be the eccentricity of the parabola {\frac  {x^{2}}{a^{2}}}-{\frac  {y^{2}}{b^{2}}}=1\,

Then b^{2}=a^{2}(e_{1}^{{2}}-1)\,

Let e2 be the eccentricity of the conjugate hyperbola

{\frac  {y^{2}}{b^{2}}}-{\frac  {x^{2}}{a^{2}}}=1\,

a^{2}=b^{2}(e_{2}^{{2}}-1)\,

Therefore

b^{2}a^{2}=a^{2}b^{2}(e_{1}^{{2}}-1)(e_{2}^{{2}}-1)\,

1=(e_{1}^{{2}}-1)(e_{2}^{{2}}-1)\,

e_{1}^{{2}}+e_{2}^{{2}}=e_{1}^{{2}}e_{2}^{{2}}\,

{\frac  {1}{e_{1}^{{2}}}}+{\frac  {1}{e_{2}^{{2}}}}=1\,


Main Page:Geometry:Hyperbola