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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 \mathrm{(0,5 \sqrt{3})}, then the length of its latus rectum is

Option: 1

3


Option: 2

5


Option: 3

4


Option: 4

6


Answers (1)

\mathrm{\text { Let equation of ellipse is } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad(a>b)}

Now, \mathrm{2 a-2 b=10 \Rightarrow a-b=5}              ....(i)

Also, \mathrm{a e=5 \sqrt{3}}                                                  ....(ii)

Now, \mathrm{a^2 e^2=a^2-b^2 \Rightarrow(5 \sqrt{3})^2=a^2-b^2}(Using (ii))

\mathrm{\Rightarrow \quad(a+b)(a-b)=75 \Rightarrow a+b=15}.(iii) (Using (i))

Solving (i) and (iii), we get \mathrm{a=10, b=5}

Length of latus rectum

\mathrm{ =\frac{2 b^2}{a}=\frac{2 \times(5)^2}{10}=\frac{50}{10}=5 }

Posted by

Ramraj Saini

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