Question

length of latus rectum of the hyperbola 25x^2-16y^2=400 is​

Answer 1

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

$\green{\tt{\therefore{Latus\:rectum(LL')=\frac{25}{2}}}}$

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

$\green {\underline \bold{Given : }} \\ \tt{ : \implies Eqn \: of \: hyperbola = 25{x}^{2}- {16y}^{2} = 400} \\ \\ \red {\underline \bold{To \: Find: }} \\ \tt {: \implies Length \: of \: latus \: rectum (LL')=?}$

• According to given question :

$\tt {: \implies {25x}^{2} - 16{y}^{2} = 400} \\ \\ \tt{: \implies \frac{25x^{2} }{400} - \frac{ 16{y}^{2} }{400} = 1 } \\ \\ \tt{ : \implies \frac{ {x}^{2} }{ \frac{400}{25} } + \frac{ {y}^{2} }{ \frac{400}{16} } = 1} \\ \\ \tt{: \implies \frac{ {x}^{2} }{16} + \frac{ {y}^{2} }{25} = 1} \\ \\ \text{So, \: it \: is \: in \: the \: form \: of} \\ \tt{\to \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \bold{Where : } \\ \tt{\circ \: {a}^{2} = 16} \\ \\ \tt{\circ \: {b}^{2} = 25} \\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies Latus \: rectum = \frac{2 {b}^{2} }{a} } \\ \\ \text{Putting \: given \: values} \\ \tt{ : \implies Latus \: rectum = \frac{2 \times 25}{4} } \\ \\ \green{\tt{ : \implies Latus \: rectum = \frac{25 }{2} }}$

Answer 2

it works well dear.....

hehe

♥️