Question

Find the vertices foci length of latus rectum and eccentricity of x^2÷16+y^2÷64=1

Answer 1

x²/16 + y²/64 = 1
(x/4)²+(y/8)² = 1

Answer 2

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

$\green{\tt{\therefore{Vertices=(0,\pm8)}}}$

$\green{\tt{\therefore{Foci=(0,\pm4\sqrt{3})}}}$

$\green{\tt{\therefore{Latus\:rectum(LL')=4}}}$

$\green{\tt{\therefore{Eccentricity=\frac{\sqrt{3}}{2}}}}$

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

$\green {\underline \bold{Given : }} \\ \tt{ : \implies eqn \: of \: ellipse = \frac{{x}^{2}}{16} + \frac{ {y}^{2} }{64} = 1} \\ \\ \red {\underline \bold{to \: find: }} \\ \tt{:\implies Vertices=?}\\ \\ \tt{:\implies Foci=?}\\ \\ \tt {: \implies Length \: of \: latus \: rectum (LL')=?}\\\\ \tt{:\implies Eccentricity=?}$

• According to given question :

$\tt{: \implies \frac{ {x}^{2} }{16} + \frac{ {y}^{2} }{64} = 1} \\ \\ \text{So, \: it \: is \: in \: the \: form \: of} \\ \tt{\to \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \bold{Where : } \\ \tt{\circ \: {a}^{2} = 16} \\ \\ \tt{\circ \: {b}^{2} = 64} \\\\ \tt{\circ\: a< b}$

$\bold{As \: we \: know \: that} \\ \tt{: \implies vertices = (0, \pm b)} \\ \\ \green{\tt{: \implies vertices = (0, \pm 8)}} \\ \\ \bold{As \: we \: know \: that} \\ \tt{ : \implies {a}^{2} = {b}^{2}(1 - {e}^{2} )} \\ \\ \tt{: \implies16 = 64(1 - {e}^{2} )} \\ \\ \tt{: \implies {e}^{2} = 1 - \frac{1}{4} } \\ \\ \tt{: \implies {e}^{2} = \frac{3}{4} } \\ \\ \green{\tt{: \implies e = \frac{ \sqrt{3} }{2} }}\\ \\ \bold{As \: we \: know \: that} \\ \tt{: \implies foci = (0, \pm be)} \\ \\ \green{\tt{: \implies Foci= (0, \pm 4 \sqrt{3} )}}$

$\bold{As \: we \: know \: that} \\ \tt{ : \implies Latus \: rectum = \frac{2 {a}^{2} }{b} } \\ \\ \text{Putting \: given \: values} \\ \tt{ : \implies Latus \: rectum = \frac{2 \times 16 }{8} } \\ \\ \green{\tt{ : \implies Latus \: rectum = 4}}$