If e and e' are eccentricities of hyperbola

**zorro** · December 4th 2009, 02:25 PM

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

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

**Soroban** · December 4th 2009, 05:25 PM

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: . $e' \:=\:\frac{c}{b}\:=\:\frac{\sqrt{a^2+b^2}}{b}$

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

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

**zorro** · December 8th 2009, 12:20 PM

Thanks u very much every one for

helping me

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Thanks u very much

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