Question

The length of the conjugate axis of a hyperbola is greater than the length of the transverse axis. Then the eccentricity e is?
$
  A) = \sqrt 2 \\
  B) > \sqrt 2 \\
  C) < \sqrt 2 \\
  D) < \dfrac{1}{{\sqrt 2 }} \\
$

Answer 1

Hint: Get the length of the conjugate axis of a hyperbola and the length of the transverse axis of hyperbola. And we know that eccentricity of hyperbola is ${b^2} = {a^2}({e^2} - 1)$. Apply the given condition.

Complete step-by-step answer:

We know that length of conjugate axis of a hyperbola = 2b

We also know that the length of transverse axis =2a

Given condition is that,

The length of conjugate axis of a hyperbola is greater than the length of transverse axis
So, by applying the given condition we can write

$
   \Rightarrow 2b > 2a \\

   \Rightarrow b > a \\

   \Rightarrow \dfrac{b}{a} > a \\
$

From the given condition we know that $\dfrac{b}{a} > 1$ from which we can also say $\dfrac{{{b^2}}}{{{a^2}}} > 1$

We also know that ${b^2} = {a^2}({e^2} - 1)$ or $\dfrac{{{b^2}}}{{{a^2}}} = {e^2} - 1$$ \to (1)$

By using the equation (1)

We can rewrite the $\dfrac{{{b^2}}}{{{a^2}}} > 1$ as

$
   \Rightarrow {e^2} - 1 > 1 \\

   \Rightarrow {e^2} > 2 \\

$

$e > \sqrt 2 $

So by applying the condition given and further simplification we got $e > \sqrt 2 $

Therefore eccentricity e is $e > \sqrt 2 $

Option B is the correct answer.

Note: In this problem we have applied the condition where we got $\dfrac{{{b^2}}}{{{a^2}}} > 1$ . To get the eccentricity of hyperbola we have used alternative value of $\dfrac{{{b^2}}}{{{a^2}}} = {e^2} - 1$