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# The eccentricity of the hyperbola whose latus rectum is equal to half of its conjugate axis is

$\begin{array}{1 1}(1)\frac{\sqrt{3}}{2}&(2)\frac{5}{3}\\(3)\frac{3}{2}&(4)\frac{\sqrt{5}}{2}\end{array}$

Given latus rectum = $\large\frac{1}{2} \times$ Conjugate axis
$\large\frac{2b^2}{a} =\frac{1}{2} \times$$2b 2b=a b^2 =a^2(e^2-1) b^2=4b^2(e^2-1) \large \frac{1}{4}=$$e^2-1$
$e^2=1+ \large\frac{1}{4}=\frac{5}{4}$
$e= \large\frac{\sqrt{5}}{2}$
Hence 4 is the correct answer.