Question:
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 the length of its latus rectum is:
Correct Option: , 2
Solution:
Given that focus is $(0,5 \sqrt{3}) \Rightarrow|b|>|a|$
Let $b>a>0$ and foci is $(0, \pm b e)$
$\because a^{2}=b^{2}-b^{2} e^{2} \Rightarrow b^{2} e^{2}=b^{2}-a^{2}$
$b e=\sqrt{b^{2}-a^{2}} \Rightarrow b^{2}-a^{2}=75$.....(i)
$\because 2 b-2 a=10 \Rightarrow b-a=5 \quad \ldots$ (ii)
From (i) and (ii)
$b+a=15$......(iii)
On solving (ii) and (iii), we get
$\Rightarrow b=10, a=5$
Now, length of latus rectum $=\frac{2 a^{2}}{b}=\frac{50}{10}=5$