If e and e’ are eccentricities of hyperbola and its conjugate res… - Vidyadip

MCQ

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

✓
${\left( {\frac{1}{e}} \right)^2} + {\left( {\frac{1}{{e'}}} \right)^2} = 1$
B
$\frac{1}{e} + \frac{1}{{e'}} = 1$
C
${\left( {\frac{1}{e}} \right)^2} + {\left( {\frac{1}{{e'}}} \right)^2} = 0$
D
$\frac{1}{e} + \frac{1}{{e'}} = 2$

✓

Answer

Correct option: A.

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

a
(a) Let hyperbola is $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$.....$(i)$

Then its conjugate will be, $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = - 1$.....$(ii)$

If $e$ is eccentricity of hyperbola $(i),$ then ${b^2} = {a^2}({e^2} - 1)$

or $\frac{1}{{{e^2}}} = \frac{{{a^2}}}{{({a^2} + {b^2})}}$ .....$(iii)$

Similarly if e' is eccentricity of conjugate $(ii),$ then ${a^2} = {b^2}(e{'^2} - 1)$ or

$\frac{1}{{e{'^2}}} = \frac{{{b^2}}}{{({a^2} + {b^2})}}$.....$(iv)$

Adding $(iii)$ and $(iv),$

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

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

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