Question

The eccentricity of the hyperbola whose latus rectum is half of its transverse axis, is
A. $\dfrac{1}{{\sqrt 2 }}$
B. $\sqrt {\dfrac{2}{3}}$
C. $\sqrt {\dfrac{3}{2}}$
D. none of these

Answer 1

The length of the latus rectum of the hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$is $\dfrac{{2{b^2}}}{a}$and transverse axis is 2a. We analyze the conditions of the problem and work according to it.

Complete step-by-step answer:
We know, the length of the latus rectum of the hyperbola $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$is $\dfrac{{2{b^2}}}{a}$and transverse axis is 2a.
And if we check, according to the question, length of latus rectum = \dfrac{1}{2}$$$$ \times length of transverse axis
Which is, $\dfrac{{2{b^2}}}{a}$= \dfrac{1}{2}$$$$ \times 2a
By cancellation process,
Or, 2${b^2} = {a^2}$
Now, eccentricity, e = $\sqrt {(1 + \dfrac{{{b^2}}}{{{a^2}}})}$= $\sqrt {(1 + \dfrac{{{b^2}}}{{2{b^2}}})}$= $\sqrt {(1 + \dfrac{1}{2})}$= $\sqrt {\dfrac{3}{2}}$
So, the eccentricity of the hyperbola is $\sqrt {\dfrac{3}{2}}$
Hence, the correct option is (C).

Note: If we rotate the axis with an angle 90 degree, then the length of the transverse axis would be = 2b and then we will find a different solution.
After rotating the axis of the hyperbola or 90 degree, we have the length of the transverse axis = 2b
And if we check, according to the question, length of latus rectum = \dfrac{1}{2}$$$$ \times length of transverse axis
Which is, $\dfrac{{2{b^2}}}{a}$= \dfrac{1}{2}$$$$ \times 2b
By cancellation process,
Or, a = 2b ,
Now, eccentricity, e = $\sqrt {(1 + \dfrac{{{b^2}}}{{{a^2}}})}$= $\sqrt {(1 + \dfrac{{{b^2}}}{{{{(2b)}^2}}})}$= $\sqrt {(1 + \dfrac{{{b^2}}}{{4{b^2}}})}$= $\sqrt {(1 + \dfrac{1}{4})}$= $\sqrt {\dfrac{5}{4}}$= $\dfrac{{\sqrt 5 }}{2}$
So, the eccentricity of the hyperbola turns out to be = $\dfrac{{\sqrt 5 }}{2}$