Question

Find the coordinates of the foci and the vertices the eccentricity and the length of the latus of rectum of the hyperbola $\frac{y^2}{9}-\frac{x^2}{27}=1$

Answer 1

The given equation is $\frac{y^2}{9}-\frac{x^2}{27}=1$ or $\frac{y^2}{3^2}-\frac{x^2}{{\left(\sqrt{27}\right)}^2}=1$
On comparing this equation with the standard equation of hyperbola i.e., $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$, we obtain $a=3$ and $b=\sqrt{27}$
We know that $a^2+b^2=c^2,$ where $c=ae$
$\therefore c^2=3^2+{\left(\sqrt{27}\right)}^2=9+27=36$
$\Rightarrow c=6$
Therefore, the coordinates of the foci are $(0,\pm 6)$
The coordinates of the vertices are $(0,\pm 3)$
Eccentricity $e=$ $\frac{c}{a}=\frac{6}{3}=2$
Length of latus rectum $=$ $\frac{2b^2}{a}=\frac{2\times 27}{3}=18$