Question

Find eccentricity foci latus rectum of x^2+4y^2+8y-2x+1=0

Answer 1

here is the answer for ur questions

Answer 2

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

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

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

$\green{\tt{\therefore{Latus\:rectum(LL')=1\:units}}}$

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

$\green {\underline \bold{Given : }} \\ \tt{ : \implies Eqn \: of \: ellipse = x^{2}+4y^{2}+8y-2x+1=0 = 1} \\ \\ \red {\underline \bold{To \: Find: }}\\ \tt{:\implies Eccentricity(e)=?}\\\\ \tt{:\implies Foci=?}\\ \\ \tt {: \implies Length \: of \: latus \: rectum (LL')=?}$

• According to given question :

$\tt {: \implies {x}^{2} + {4y}^{2} + 8y - 2x + 1 = 0} \\ \\ \tt{: \implies {x}^{2} - 2x + 1 - 1 + 4 {y}^{2} + 8y + 4 -4+ 1 = 0} \\ \\ \tt{: \implies (x - 1)^{2} +( 2y + 2)^{2} = 4} \\ \\ \tt{ :\implies \frac{(x - 1)^{2} }{4} + \frac{(y + 1) ^{2} }{ \frac{4}{4} } = 1 } \\ \\ {: \implies \frac{ {(x - 1)}^{2} }{4} + \frac{ {(y + 1)}^{2} }{1} = 1}$

$\text{So, \: it \: is \: in \: the \: form \: of} \\ \tt{\to \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1} \\ \\ \bold{Where : } \\ \tt{\circ \: {a}^{2} = 4} \\ \\ \tt{\circ \: {b}^{2} = 1} \\\\ \tt{\circ\: a > b}$

$\bold{As \: we \: know \: that} \\ \tt{ : \implies {b}^{2} = {a}^{2}(1 - {e}^{2} )} \\ \\ \tt{: \implies1 = 4(1 - {e}^{2} )} \\ \\ \tt{: \implies {e}^{2} = 1 - \frac{1}{4} } \\ \\ \tt{: \implies {e}^{2} = \frac{3}{4} } \\ \\ \green{\tt{: \implies e = \frac{ \sqrt{3} }{2} }}\\ \\ \bold{As \: we \: know \: that} \\ \tt{: \implies foci = ( \pm ae,0)} \\ \\ \green{\tt{: \implies Foci= (\pm \sqrt{3}, 0)}}$

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