Question

Find the length of transverse axis, length of
conjugate axis, the eccentricity, the co-ordi-
nates of foci, equations of directrices and the
length of latus rectum of the hyperbola.
x² ÷25 - y^2 = 1

​

Answer 1

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

$\green{\tt{\therefore{Length\:of\:transverse\:axis=10\:units}}}$

$\green{\tt{\therefore{Length\:of\:conjugate\:axis=2\:units}}}$

$\green{\tt{\therefore{Foci=(\pm\sqrt{30},0)}}}$

$\green{\tt{\therefore{Latus\:rectum(LL')=0.4\:units}}}$

$\green{\tt{\therefore{Eqn\:of\:directrix=x\pm5\sqrt{5}}}}$

$\green{\tt{\therefore{Eccentricity=\sqrt{\frac{6}{5}}}}}$

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

$\green {\underline \bold{Given : }} \\ \tt{ : \implies Eqn \: of \: hyperbola = \frac{{x}^{2}}{25} -\frac{ {y}^{2} }{1} = 1} \\ \\ \red {\underline \bold{to \: find: }}\\ \tt{:\implies Length\:of\:transverse\:axis=?} \\ \\\tt{:\implies Length\:of\:conjugate\:axis=?}\\\\ \tt{:\implies Vertices=?}\\ \\ \tt{:\implies Foci=?}\\ \\ \tt {: \implies Length \: of \: latus \: rectum (LL')=?}\\\\ \tt{:\implies Eccentricity=?}\\\\ \tt{:\implies Eqn\:of\:directrix=?}$

• According to given question :

$\tt{: \implies \frac{ {x}^{2} }{25} - \frac{ {y}^{2} }{1} = 1} \\ \\ \text{So, \: it \: is \: in \: the \: form \: of} \\ \tt{\to \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \bold{Where : } \\ \tt{\circ \: {a}^{2} = 25} \\ \\ \tt{\circ \: {b}^{2} = 1}$

$\bold{as \: we \: know \: that} \\ \tt{: \implies length \: of \: transverse \: axis = 2a} \\ \\ \tt{: \implies length \: of \: transverse \: axis = 2 \times 5} \\ \\ \green{\tt{: \implies length \: of \: transverse \: axis = 10 \: units}} \\ \\ \bold{as \: we \: know \: that} \\ \tt{: \implies length \: of \:conjugate \: axis = 2b} \\ \\ \tt{: \implies length \: of \: conjugate \: axis = 2 \times 1} \\ \\ \green{\tt{: \implies length \: of \: conjugate \: axis = 2\: units}}$

$\bold{As \: we \: know \: that} \\ \tt{ : \implies {b}^{2} = {a}^{2}({e}^{2} -1 )} \\ \\ \tt{: \implies1= 5( {e}^{2} -1)} \\ \\ \tt{: \implies {e}^{2} = 1 + \frac{1}{5} } \\ \\ \tt{: \implies {e}^{2} = \frac{6}{5} } \\ \\ \green{\tt{: \implies e = \sqrt{\frac{6}{5} }}}\\ \\ \bold{As \: we \: know \: that} \\ \tt{: \implies foci = (\pm ae,0)} \\ \\ \green{\tt{: \implies Foci= (\pm \sqrt{30} ,0)}}$

$\bold{as \: we \: know \: that} \\ \tt{: \implies eqn \: of \: directrix = x = \pm\frac{a}{e} } \\ \\ \tt{: \implies x = \pm \frac{5}{ \frac{1}{ \sqrt{5} } } } \\ \\ \tt {: \implies x = \pm 5 \sqrt{5} } \\ \\ \green{\tt {: \implies x \pm5 \sqrt{5} = 0}}$

$\bold{As \: we \: know \: that} \\ \tt{ : \implies Latus \: rectum = \frac{2 {b}^{2} }{a} } \\ \\ \text{Putting \: given \: values} \\ \tt{ : \implies Latus \: rectum = \frac{2 \times 1}{5} } \\ \\ \green{\tt{ : \implies Latus \: rectum = 0.4\:units}}$