Question

Find the equation of the hyperbola the distance whose foci is 16 and whose eccentricity is√2

Answer 1

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

$\green{\tt{\therefore{Eqn\:of\:hyperbola=x^{2}-y^{2}=32}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies Foci = 16} \\ \\ \tt{ : \implies Eccentricity(e) = \sqrt{2}} \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{ : \implies Eqn \:of \: hyperbola = ?}$

• According to given question :

$\circ \: \tt{Let \: eqn \: be \: \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} - - - - - (1) \\ \\ \ \tt{: \implies Foci = 2ae} \\ \\ \tt{ : \implies Foci \ = 16} \\ \\ \tt{ : \implies 2ae =16} \\ \\ \tt{: \implies a \times \sqrt{2} = 8} \\ \\ \green{ \tt{: \implies a = 4\sqrt{2}} }\\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies {b}^{2} = {a}^{2} ( {e}^{2} - 1)} \\ \\ \tt{: \implies {b}^{2} = ({4\sqrt{2}})^{2} ( {\sqrt{2}}^{2} - 1)} \\ \\ \tt{ : \implies {b}^{2} = 32(2 - 1)} \\ \\ \tt{: \implies {b}^{2} = 32} \\ \\ \text{Putting \: given \: values \: in \: (1)} \\ \tt{ : \implies \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \green{ \tt{: \implies \frac{ {x}^{2} }{32} - \frac{ {y}^{2} }{32} = 1}} \\ \\ \green{\tt{\therefore Eqn \: of \:hyperbola \: is \: {x}^{2} - {y}^{2} = 32}}$