Question

find the eccentricity foci equation of directrix of the hyperbola 7x²-8y²=11

Answer 1

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

$\green{\tt{\therefore{Eccentricity=\sqrt{\frac{15}{8}}}}}$

$\green{\tt{\therefore{Foci=\pm(\sqrt{\frac{15}{8}}\times \sqrt{\frac{11}{7}},0)}}}$

$\green{\tt{\therefore{Eqn\:of\:directrix=x\pm\sqrt{\frac{88}{105}}=0}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies Eqn\:of\:hyperbola = 7x^{2}-8y^{2}=11} \\ \\ \red{ \underline \bold{To \: Find : }}\\ \tt{ : \implies Eccentricity(e) = ?} \\ \\ \tt{ : \implies Foci = ?}\\\\ \tt{:\implies Eqn\:of\:directrix=?}$

• According to given question :

$\tt{ : \implies {7 x }^{2} - {8y}^{2} = 11} \\ \\ \tt{: \implies \frac{ {x}^{2} }{ \frac{11}{7} } - \frac{ {y}^{2} }{ \frac{11}{8} } = 1} \\ \\ \text{so \: it \: is \: in \: the \: form \: of }\\ \tt{ \to \: \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \circ \: \tt{{a}^{2} = \frac{11}{7} } \\ \\ \circ \: \tt{ {b}^{2} = \frac{11}{8} } \\ \\ \bold{as \: we \: know \: that } \\ \tt{: \implies {b}^{2} = {a}^{2} ( {e}^{2} - 1)} \\ \\ \tt{\implies \frac{11}{8} = \frac{11}{7} ( {e}^{2} - 1)} \\ \\ \tt{ : \implies \frac{7}{8} = {e}^{2} - 1} \\ \\ \tt {: \implies {e}^{2} = \frac{7}{8} + 1} \\ \\ \green{\tt{: \implies e = \sqrt{ \frac{15}{8} } }} \\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies foci = ( \pm ae,0)} \\ \\ \green{\tt{ : \implies foci = ( \pm\sqrt{ \frac{11}{7} } \times \sqrt{\frac{15}{8}} ,0})} \\ \\ \bold{As \: we \: know \: that} \\ \tt{: \implies eqn \: of \: directrix = x = \pm\frac{a}{e}} \\ \\ \tt{ : \implies x = \pm\sqrt{\frac{ \frac{11}{7} }{\frac{15}{8}} }} \\ \\ \green{ \tt {: \implies x \pm \sqrt{ \frac{88}{105} } = 0}}$