Question

find the eccentricity and foci of the hyperbola 9y^2-4x^2=36.​

Answer 1

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

$\green{\tt{\therefore{Eccentricity=\frac{\sqrt{13}}{2}}}}$

$\green{\tt{\therefore{Foci=(0,\pm\sqrt{15})}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies Eqn\:of\:hyperbola = 4x^{2}-9y^{2}=-36} \\ \\ \red{ \underline \bold{To \: Find : }}\\ \tt{ : \implies Eccentricity(e) = ?} \\ \\ \tt{ : \implies Foci = ?}$

• According to given question :

$\tt {: \implies -{4x}^{2} + 9 {y}^{2} = 36} \\ \\ \tt{: \implies -\frac{4x^{2} }{36} + \frac{ 9{y}^{2} }{36} = 1 } \\ \\ \tt{ : \implies -\frac{ {x}^{2} }{ \frac{36}{4} } + \frac{ {y}^{2} }{ \frac{36}{9} } = 1} \\ \\ \tt{: \implies -\frac{ {x}^{2} }{9} + \frac{ {y}^{2} }{4} = 1} \\ \\ \text{So, \: it \: is \: in \: the \: form \: of} \\ \tt{\to -\frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \bold{Where : } \\ \tt{\circ \: {a}^{2} = 9} \\ \\ \tt{\circ \: {b}^{2} = 4} \\ \\ \bold{As \: we \: know \: that}$

$\tt{: \implies {a}^{2} = {b}^{2} ( {e}^{2} - 1)} \\ \\ \tt{:\implies 9 = 4( {e}^{2} - 1)} \\ \\ \tt{ : \implies \frac{9}{4} = {e}^{2} - 1} \\ \\ \tt {: \implies {e}^{2} = \frac{9}{4} + 1} \\ \\ \green{\tt{:\implies e = \frac{\sqrt{13}}{2} }} \\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies foci = (0,\pm be)} \\ \\ \tt{ : \implies foci = (0, \pm 2\times \frac{\sqrt{13}}{2} )} \\ \\ \green{ \tt {: \implies Focus = (0,\pm \sqrt{13})}}$