in an ellipse with centre at the origin , if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at `(0,5,sqrt(3))`, then length of its latus rectum is

Options

A. 5
B. 10
C. 8
D. 6

Answer

A

Explanation

one of the focus of ellipse `(x^(2))/(a^(2)) +(y^(2))/(b^(2))=1` is on Y-axis ` (0,5,sqrt(3) )`
`therefore be = 5 sqrt(3) `
[ where e is eccentricity of ellipse ]
According to the question ,
`2b-2a=10`
` implies b-a=5`
on squaring Eq,. (i) both sides , we get
`b^(2) e^(2) =75`
`implies b^(2) (1-(a^(2))/(b^(2)))=75`
`implies b^(2) -a^(2)=75[:' e^(2)=1-(a^(2))/(b^(2))]`
`implies b^(2) -a^(2)=75`
`implies (b+a)(b-a)=75`
`implies b+a=15`
`implies b+a+15`[ from Eq. (ii) ] . . . (iii)
on solving Eqs. (ii) and (iii) we get
So , length of Latusrectum is `(2a^(2))/(b)=(2xx25)/(10)= 5` units