Find the equation of the ellipse coordinates of whose foci are (±2, 0) and length of latus rectum is 10 3 .

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Foci are S2,0 and S'-2,0.As Foci lie on X axis, hence equation of ellipse will bex2a2+y2b2=1 a>bNow foci for above ellipse are  Sae,0  and S'-ae,0 ae=2e=2a ;iLength of ratus ractum=2b2a=10b2=5a ;iia21-e2=5aa1-e2=5a1-4a2=5aa2-4a2=5a2-4=5aa2-5a-4=0a=--5±52-41-42=5±412But a>0a=5+412 ;iiia2=5+4124=25+41+10414=66+10414=33+5412Using ii and iii, we haveb2=5.5+412=25+5412Hence equation of ellipse is:x233+5412+y225+5412=1

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