If the latus rectum of a hyperbola subtend  an angle of 60° at the other focus, then eccentricity of the hyperbola is

Let the equation of the hyperbola be x2a2−y2b2=1 F1(−ae,0),F2(ae,0) be the foci, e being the eccentricity.Let LF2L′, the latus rectum subtend an angle of 60° at F1, then slope of LF1 is tan 300.     So tan⁡ 30∘=b2/a2ae⇒ 13=a2e2−12a2e=e2−12e⇒ e2−1=23e⇒e−132=1+13⇒ e−13=23⇒e=3