Math, asked by Rizwee, 1 year ago

find the eccentricity of the hyperbola whose latus rectum and transverse axis are of same lenght​

Answers

Answered by BrainlyConqueror0901
1

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

\green{\tt{\therefore{Eccentricity=\sqrt{2}}}}

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

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

• According to given question :

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

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