Question

Find the eccentricity and length of latus rectum of hyperbola
$\frac{x^2}{25}-\frac{y^2}{9}=1$

Answer 1

The equation of the hyperbola is
$\frac{x^2}{25}-\frac{y^2}{9}=1$
There, $a^2=25\Rightarrow a=5$
and $b^2=9\Rightarrow b=3$
We know $b^2=a^2(e^2-1)$
$\Rightarrow 9=25(e^2-1)$
$\Rightarrow e^2-1=\frac{9}{25}$
$\Rightarrow e^2=\frac{9}{25}+1=\frac{34}{25}$
$\Rightarrow e=\sqrt{\frac{34}{25}}=\frac{\sqrt{34}}{5}$
Length of latus rectum $=\frac{2b^2}{a}$
$=2\times \frac{9}{5}=\frac{18}{5}$ units