Question

The ecentricity of hyperbola is 2 and its Length of foci is 12.Then find the eqn of hyperbola.​

Answer 1

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Answer 2

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

$\green{\tt{\therefore{Eqn\:of\:hyperbola=3x^{2}-y^{2}=27}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies Foci = 12} \\ \\ \tt{ : \implies Eccentricity(e) = 2} \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{ : \implies Eqn \:of \: hyperbola = ?}$

• According to given question :

$\circ \: \tt{Let \: eqn \: be \: \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} - - - - - (1) \\ \\ \ \tt{: \implies Foci = 2ae } \\ \\ \tt{ : \implies 12 =2ae } \\ \\ \tt{: \implies a \times {2} = 6} \\ \\ \green{ \tt{: \implies a = 3 }}\\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies {b}^{2} = {a}^{2} ( {e}^{2} - 1)} \\ \\ \tt{: \implies {b}^{2} = 3^{2}( 2^{2} - 1)} \\ \\ \green{ \tt{ : \implies {b}^{2} =9\times 3} }\\ \\ \tt{: \implies {b}^{2} = 27}\\\\ \text{Putting \: given \: values \: in \: (1)} \\ \tt{ : \implies \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \green{ \tt{: \implies \frac{ {x}^{2} }{9} - \frac{ {y}^{2} }{27} = 1}} \\ \\ \green{\tt{\therefore Eqn \: of \:hyperbola \: is \: 3{x}^{2} - {y}^{2} = 27}}$