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The asymptotes of a hyperbola all satisfy x² - 4y² = 0. Find the possible value(s) of the eccentricity of such a hyperbola. (Multiple correct)
- sqrt(5)/2
- sqrt(5)
- sqrt(5)/3
- 3/2
Correct answer: sqrt(5)/2
Solution
If b/a = 1/2 then e = sqrt(1 + b²/a²) = sqrt(1 + 1/4) = sqrt5/2; the conjugate (vertical) hyperbola with a/b = 1/2 gives e = sqrt(1 + 4) = sqrt5, so both sqrt5/2 and sqrt5 are valid (the principal accepted value being sqrt5/2).
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