Question 933264


the equation of ellipse whose vertices are ({{{5}}},{{{0}}}) and ({{{-5}}},{{{0}}}) and length of latus rectum is {{{8/5}}}:

standard form of equation: 

{{{x^2/a^2+y^2/b^2=1}}}, ....{{{a>b}}}

vertices:({{{0}}}±{{{a}}},{{{0}}})=({{{5}}},{{{0}}}) => {{{a=5}}} and {{{a=-5}}}

The Latus Rectum is the line through the focus and parallel to the directrix.

The length of the Latus Rectum is {{{2b^2/a}}}

since latus rectum is {{{8/5}}} and {{{a=5}}}, we have

{{{2b^2/5= 8/5}}} now find {{{b}}},..denominators same then

{{{2b^2= 8}}}

{{{b^2= 4}}}

{{{b=2}}} or {{{b=-2}}}

 {{{x^2/5^2+y^2/2^2=1}}}

 {{{x^2/25+y^2/4=1}}}

{{{ graph( 600, 600, -10, 10, -10, 10,-sqrt(4-4x^2/25) ,sqrt(4-4x^2/25)) }}}