The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distance between the foci, is

For the standard hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$ the length of the latus rectum is $$\frac{2b^2}{a}=8.$$ So $$b^2=4a.$$ The conjugate axis has length $$2b,$$ and the distance between the foci is $$2c.$$ Given that the conjugate axis is half the distance between the foci, $$2b=\frac{1}{2}(2c),$$ hence $$c=2b.$$ Using $$c^2=a^2+b^2,$$ we get $$4b^2=a^2+b^2.$$ So $$a^2=3b^2.$$ Therefore $$a=\sqrt{3}b.$$ The eccentricity is $$e=\frac{c}{a}=\frac{2b}{\sqrt{3}b}=\frac{2}{\sqrt{3}}.$$ Comparing with the options, this matches option $C$.