Question

Find length of latus rectum and eccentricity of an ellipse whose equation is 4x^2+6y^2=24.​

Answer 1

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

$\green{\tt{\therefore{Latus\:rectum=\frac{8}{\sqrt{6}}}}}$

$\green{\tt{\therefore{Eccentricity=\frac{1}{\sqrt{3}}}}}$

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

$\green{\underline \bold{Given:}} \\ \tt: \implies Eqn \: of \: ellipse = {4x}^{2} + 6 {y}^{2} = 24 \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Latus \: rectum = ? \\ \\ \tt: \implies Eccentricity = ?$

• According to given question :

$\tt: \implies 4 {x}^{2} + {6y}^{2} = 24 \\ \\ \tt: \implies \frac{ 4{x}^{2} }{24} + \frac{6 {y}^{2} }{24} = 1 \\ \\ \tt: \implies \frac{ {x}^{2} }{6} + \frac{ {y}^{2} }{4} = 1 \\ \\ \text{So, \: it \: is \: in \: the \: form \: of \: } \\ \tt: \implies \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1 \\ \\ \bold{Where : } \\ \tt{\circ \: {a}^{2} = 6} \\ \\ \tt{\circ \: {b}^{2} = 4} \\ \\ \bold{As \: we \: know \: that}$

$\tt: \implies Latus \: rectum = \frac{2 {b}^{2} }{a} \\ \\ \tt: \implies Latus \: rectum = \frac{2 \times 4}{ \sqrt{6} } \\ \\ \green{\tt: \implies Latus \: rectum = \frac{8}{ \sqrt{6} \:} } \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {b}^{2} = {a}^{2} (1 - {e}^{2} ) \\ \\ \tt: \implies 4 = 6(1 - {e}^{2} ) \\ \\ \tt: \implies \frac{4}{6} = 1 - {e}^{2} \\ \\ \tt: \implies {e}^{2} = 1 - \frac{4}{6} \\ \\ \green{\tt: \implies e = \frac{1}{\sqrt{3} } }$

Answer 2

$\mathfrak{\huge{\pink{ANSWER}}}$