conic-section

Question: If e and e’ are eccentricities of hyperbola and its conjugate respectively, then

1) ${{\left( \frac{1}{e} \right)}^{2}}+{{\left( \frac{1}{e'} \right)}^{2}}=1$

2) $\frac{1}{e}+\frac{1}{e'}=1$

3) ${{\left( \frac{1}{e} \right)}^{2}}+{{\left( \frac{1}{e'} \right)}^{2}}=0$

4) $\frac{1}{e}+\frac{1}{e'}=2$

Solution:

Explanation: No Explanation

Hyperbola

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