Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$
The given equation is $\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$ or $\frac{y^{2}}{3^{2}}-\frac{x^{2}}{(\sqrt{27})^{2}}=1$
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 known that $a^{2}=b^{2}+c^{2}$
$\therefore c^{2}=3^{2}+(\sqrt{27})^{2}=9+27=36$
$\Rightarrow c=6$
The coordinates of the foci are $(0,\,±6)$
The coordinates of the vertices are $(0,\,±3)$
Eccentricity, $e=\frac{c}{a}=\frac{6}{3}=2$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 27}{3}=18$
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