Question

find equation of ellipse reffered to its curve whose latus rectum is 5 and eccentricity is 2/3

Answer 1

hope it will help u out......

Answer 2

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

$\green{\tt{\therefore{Eqn\:of\:ellipse=\frac{x^{2}}{20.25}+\frac{y^{2}}{11.25}=1}}}$

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

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies Eccentricity = \frac{2}{3}} \\ \\ \tt{ : \implies Length\:of\:latus\:rectum= 5} \\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{ : \implies Eqn \:of \: hyperbola = ?}$

• According to given question :

$\circ \: \tt{Let \: eqn \: be \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1} - - - - - (1) \\ \\ \bold{As\:we\:know\:that}\\ \tt{:\implies Length\:of\:latus\:rectum=\frac{2b^{2}}{a}}\\\\ \tt{:\implies 5=\frac{2b^{2}}{a}}\\\\ \tt{:\implies b^{2}=\frac{5a}{2}-----(2)}\\\\ \bold{As \: we \: know \: that} \\ \tt{ : \implies {b}^{2} = {a}^{2} (1- {e}^{2})} \\ \\ \tt{: \implies \frac{5a}{2} = a^{2} (1-(\frac{2}{3})^{2})} \\ \\ \tt{ : \implies \frac{5}{2a} = \frac{5}{9} }\\\\ \green{\tt{:\implies a=4.5}}$

$\text{putting \: value \: of \: a \: in \: (2)} \\ \tt{: \implies {b}^{2} = \frac{5 \times 4.5}{2} } \\ \\ \tt{: \implies {b}^{2} = 11.25} \\ \\ \text{putting \: value \: of \: a \: and \: b \: in \: (1)} \\ \tt{ : \implies \frac{ {x}^{2} }{20.25} + \frac{ {y}^{2} }{11.25} = 1} \\ \\ \green{\tt{\therefore Eqn \: of \: ellipse \: is \: \frac{ {x}^{2} }{20.25} + \frac{ {y}^{2} }{11.25} = 1}}$