Geo5.2.11

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Find the equation of ellipse with length of latus rectum 15/2 and distance between foci 2.

Let the equation of the ellipse be {\frac  {x^{2}}{a^{2}}}+{\frac  {y^{2}}{b^{2}}}=1\,

Given that

{\frac  {2b^{2}}{a}}={\frac  {15}{2}},4b^{2}=15a\,

2ae=2,ae=1\,

b^{2}=a^{2}(1-e^{2}),15a=4a^{2}(1-e^{2}),15a=4a^{2}-4,4a^{2}-15a-4=0\,

4a^{2}-16a+a-4=0,(a-4)(4a+1)=0,a=4\,

4b^{2}=60,b^{2}=15\,

The equaion of the ellipse is

15x^{2}+16y^{2}=240\,

Main Page:Geometry:The Ellipse