Question

The hyperbola has its conjugate axis 5 and passes through the point (2, –1). the length of a latus-rectum is

Answer 1

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

$\green{\tt{\therefore{Latus\:rectum(LL')=1.25\sqrt{29}}}}$

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

$\green {\underline \bold{Given : }} \\ \tt{ : \implies Conjugate \: axis = 5} \\ \\ \tt{:\implies Point\:of\:hyperbola=(2,-1)} \\ \\ \red {\underline \bold{To \: Find: }} \\ \tt {: \implies Length \: of \: latus \: rectum (LL')=?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Conjugate \: axis = 2b} \\ \\ \tt{: \implies 5= 2b} \\ \\ \green{\tt{: \implies b = \frac{5}{2} }} \\ \\ \bold{Eqn \: of \: standard \: hyperbola} \\ \tt{: \implies \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1 } \\ \\ \text{Putting \: value \: of \: b} \\ \tt{: \implies \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {( \frac{5}{2} })^{2} } = 1} \\ \\ \tt{: \implies \frac{ {x}^{2} }{ {a}^{2} } - \frac{ 4{y}^{2} }{25} = 1} \\ \\ \text{(2, - 1) \: satisfy \: on \: this \: eqn} \\ \tt{: \implies \frac{ {2}^{2} }{ {a}^{2} } - \frac{4 {( - 1)}^{2} }{25} = 1} \\ \\ \tt{: \implies \frac{4}{ {a}^{2} } - \frac{4}{25} = 1} \\ \\ \tt{: \implies \frac{4}{ {a}^{2} } = 1 + \frac{4}{25} } \\ \\ \tt{: \implies \frac{4}{ {a}^{2} } = \frac{29}{25} } \\ \\ \tt{: \implies 29 {a}^{2} = 100} \\ \\ \tt{: \implies {a}^{2} = \frac{100}{29} } \\ \\ \bold{As \: we \: know \: that} \\ \tt{: \implies latus \: rectum = \frac{2 {b}^{2} }{a} } \\ \\ \tt{: \implies latus \: rectum = \frac{2 \times \frac{25}{4} }{ \frac{10}{ \sqrt{29} } } } \\ \\ \green{\tt{: \implies latus \: rectum = 1.25 \sqrt{29}} }$