Question

Find ecentricity of hyperbola whose transverse axis is half half of its latus rectum.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Eccentricity=\sqrt{3}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \tt{: \implies Transverse \: axis \: = half \: of \: {Latus \: rectum} } \\ \\ \red{ \underline \bold{To \: Find: }} \\ \tt{: \implies Eccentricity(e) =?}$

• According to given question :

$\tt{: \implies Transverse \: axis = half \: of \: latus \: rectum} \\ \\ \bold{As \: we \: know \: that} \\ \tt{: \implies 2a = \frac{ \frac{2 {b}^{2} }{a} }{2}} \\ \\ \tt{: \implies 2a = \frac{2 {b}^{2} }{2a} } \\ \\ \tt{: \implies 2 {a}^{2} = {b}^{2} } - - - - - (1) \\ \\ \bold{As \: we \: know \:that} \\ \tt{: \implies {b}^{2} = {a}^{2} ( {e}^{2} - 1)} \\ \\ \text{Putting \: value \: of \: {b}}^{2}\\ \tt{: \implies {2a}^{2} = {a}^{2} ( {e}^{2} - 1)} \\ \\ \tt{: \implies 2{a}^{2} = {a}^{2} ( {e}^{2} - 1)} \\ \\ \tt{: \implies {e}^{2} - 1 = 2} \\ \\ \tt{: \implies {e}^{2} = 3} \\ \\ \green{\tt{: \implies e = \sqrt{3} }}$