Question

The lenth of latus rectum of hyperbola (X)2÷50-(y)2÷25=1 is?

Answer 1

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

$\green{\tt{\therefore{Latus\:rectum(LL')=\sqrt{50}}}}$

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

$\green {\underline \bold{Given : }} \\ \tt{ : \implies Eqn \: of \: hyperbola = \frac{{x}^{2}}{50}- \frac{{y}^{2}}{25} = 1} \\ \\ \red {\underline \bold{To \: Find: }} \\ \tt {: \implies Length \: of \: latus \: rectum (LL')=?}$

• According to given question :

$\tt{: \implies \frac{ {x}^{2} }{50} + \frac{ {y}^{2} }{25} = 1} \\ \\ \text{So, \: it \: is \: in \: the \: form \: of} \\ \tt{\to \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \bold{Where : } \\ \tt{\circ \: {a}^{2} = 50} \\ \\ \tt{\circ \: {b}^{2} = 25} \\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies Latus \: rectum = \frac{2 {b}^{2} }{a} } \\ \\ \text{Putting \: given \: values} \\ \tt{ : \implies Latus \: rectum = \frac{2 \times 25}{\sqrt{50}} } \\ \\ \green{\tt{ : \implies Latus \: rectum = \sqrt{50} }}$