Question

If the length of the latus rectum of ellipse is 5/2 and eccentricity is ½. Then the equation of the ellipse in standard form is

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

$\tt \: Let \: the \: equation \: of \: ellipse \: be \: \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

☆ According to statement,

$\tt \longrightarrow \: eccentricity \: (e) = \dfrac{1}{2}$

☆ As we know,

$\tt \longrightarrow \: {b}^{2} = {a}^{2}(1 - {e}^{2} ) \\ \tt \longrightarrow \: {b}^{2} = {a}^{2} \bigg( 1 -\dfrac{1}{4} \bigg) \\ \tt \longrightarrow \: {b}^{2} = \dfrac{3}{4} {a}^{2} \: - - (i)$

☆ Also, it is given that

$\tt \longrightarrow \: Length \: of \: Latus \: Rectum \: = \dfrac{5}{2}$

$\tt\implies \:\dfrac{ {2b}^{2} }{a} = \dfrac{5}{2}$

$\tt\implies \: {b}^{2} = \dfrac{5}{4} a - - - (ii)$

☆ On equating (i) and (ii), we get

$\tt \longrightarrow \: \dfrac{3}{4} {a}^{2} = \dfrac{5}{4} a$

$\tt\implies \boxed{ \red{ \bf \: \:a \: = \dfrac{5}{3} }}$

☆ On substituting value of a in equation (ii), we get

$\tt \longrightarrow \: {b}^{2} = \dfrac{5}{4} \times \dfrac{5}{3}$

$\tt\implies \: \boxed{ \red{ \bf \: {b}^{2} = \dfrac{25}{12} }}$

☆ So required equation of ellipse is

$\tt \longrightarrow \: \dfrac{ {x}^{2} }{\dfrac{ 25 }{9} } + \dfrac{ {y}^{2} }{\dfrac{25}{12} } = 1 \\ \\ \tt \longrightarrow \: \dfrac{ {9x}^{2} }{25} + \dfrac{ {12y}^{2} }{25} = 1 \\ \\ \tt \longrightarrow \: {9x}^{2} + {12y}^{2} = 25$