Question

what is the length of latus rectum of ellipse 8x^2-y^2=64

Answer 1

COORECT QUESTION :

what is the length of latus rectum of hyperbola 8x^2-y^2=64.

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

$\green{\tt{\therefore{Latus\:rectum(LL')=2\:units}}}$

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

$\green {\underline \bold{Given : }} \\ \tt{ : \implies Eqn \: of \: hyperbola = 8{x}^{2}- {y}^{2} = 64} \\ \\ \red {\underline \bold{To \: Find: }} \\ \tt {: \implies Length \: of \: latus \: rectum (LL')=?}$

• According to given question :

$\tt {: \implies {8x}^{2} - {y}^{2} = 64} \\ \\ \tt{: \implies \frac{8x^{2} }{64} - \frac{ {y}^{2} }{64} = 1 } \\ \\ \tt{ : \implies \frac{ {x}^{2} }{ \frac{64}{8} } + \frac{ {y}^{2} }{ \frac{64}{1} } = 1} \\ \\ \tt{: \implies \frac{ {x}^{2} }{8} + \frac{ {y}^{2} }{64} = 1} \\ \\ \text{So, \: it \: is \: in \: the \: form \: of} \\ \tt{\to \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \bold{Where : } \\ \tt{\circ \: {a}^{2} = 8} \\ \\ \tt{\circ \: {b}^{2} = 64} \\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies Latus \: rectum = \frac{2 {a}^{2} }{b} } \\ \\ \text{Putting \: given \: values} \\ \tt{ : \implies Latus \: rectum = \frac{2 \times 8}{8} } \\ \\ \green{\tt{ : \implies Latus \: rectum = 2\:units}}$