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The ends of the major axis of an ellipse are (- 2, 4) and (2, 1). If the point (1, 3) lies on the ellipse,then find its latus rectum and eccentricity.

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the major end points are (-2,4),(2,1)
by distance formula
`2a=sqrt(4^2+3^2)`
`2a=sqrt(16+9)`
`2a=5`
`a=5/2`
center of eclipse `((-2+2)/2,(4+1)/2)`
equation of eclipse
`(x-0)^2/a^2+(y-(5/2))^2/b^2=1`
`x^2/(25/4)+(y-(5/2))^2/b^2=1`
now, we have to find b
putting the value(1,3)
`4/25+1/4b_2=1`
`1/4b^2=1-4/5=21/25`
so `b^2=25/84`
`x^2/(25/4)+(y-(5/2))^2/b62=1`
length of latus rectum will be`=2b^2/a=(2*25*2)/(84*5)=5/21`
eccentricity=`e=sqrt(1-b^2/a^2)=sqrt(1-(25/84)/(25/4)=sqrt(20/21)`.

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