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

Hint: 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 = \[\dfrac{{\sqrt 5 }}{2}\]