Question

If the eccentricity of an ellipse be $\frac{5}{8}$ and the distance between its foci be 10, then its latus rectum is

• A. 39/4
• B. 12
• C. 15
• D. 37/2

Answer 1

The correct option is A $(\frac{39}{4}$)

Given: the distance between its foci be 10 and is denoted by
$2c=10\Rightarrow c=5$.
As we know that the eccentricity ,$e=\frac{c}{a}$
$\Rightarrow \frac{5}{8}=\frac{5}{a}$
$\therefore a=8$
Also $c^2=a^2-b^2$
$\Rightarrow b^2=a^2-c^2$
$=8^2-5^2$
$=64-25$
$\therefore b=39$
Thus, length of the latus rectum is $=\frac{2b^2}{a}$
$=\frac{2(39{)}^2}{8}$
$=\frac{39}{4}$