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Math Help - If e and e' are eccentricities of hyperbola

  1. #1
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    If e and e' are eccentricities of hyperbola

    Question :
    If e and e' are eccentricities of hyperbola and its conjugate, prove that

    <br />
\left(\frac{1}{e^2} \right) + \left( \frac{1}{e'^2} \right) = 1<br />
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  2. #2
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    Hello, zorro!

    If e and e' are eccentricities of a hyperbola and its conjugate,

    prove that: . \frac{1}{e^2}+ \frac{1}{(e')^2} \:=\: 1

    A hyperbola is: . \frac{x^2}{a^2} - \frac{y^2}{b^2} \:=\:1
    . . Its eccentricity is: . e \:=\:\frac{c}{a}\:=\:\frac{\sqrt{a^2+b^2}}{a}

    Its conjugate is: . \frac{y^2}{b^2} - \frac{x^2}{a^2} \:=\:1
    . . Its eccentricity is: . e' \:=\:\frac{c}{b}\:=\:\frac{\sqrt{a^2+b^2}}{b}


    \text{Hence: }\;\frac{1}{e^2} + \frac{1}{(e')^2} \;\;=\;\;\frac{1}{\left(\frac{\sqrt{a^2+b^2}}{a}\r  ight)^2} + \frac{1}{\left(\frac{\sqrt{a^2+b^2}}{b}\right)^2}

    . . . =\;\;\frac{1}{\frac{a^2+b^2}{a^2}} + \frac{1}{\frac{a^2+b^2}{b^2}} \;\;=\;\; \frac{a^2}{a^2+b^2} + \frac{b^2}{a^2+b^2} \;=\;\frac{a^2+b^2}{a^2+b^2} \;\;=\;\;1

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  3. #3
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    Thanks u very much

    Thanks u very much every one for helping me
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