# Geo5.3.45

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

$b^2 a^2=a^2 b^2(e_1^{2}-1)(e_2^{2}-1)\,$

$1=(e_1^{2}-1)(e_2^{2}-1)\,$

$e_1^{2}+e_2^{2}=e_1^{2} e_2^{2}\,$

$\frac{1}{e_1^{2}}+\frac{1}{e_2^{2}}=1\,$

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