Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
The given equation is $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$ or $\frac{x^{2}}{4^{2}}-\frac{y^{2}}{3^{2}}=1$
On comparing this equation with the standard equation of hyperbola i.e., $\frac{ x ^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,$ we obtain $a=4$ and $b=3$.
We known that $a^{2}=b^{2}+c^{2}$
$\therefore c^{2}=4^{2}+3^{2}=25$
$\Rightarrow c=5$
Therefore,
The coordinates of the foci are $(±5,\,0)$
The coordinates of the vertices are $(±4,\,0)$
Eccentricity, $e=\frac{c}{a}=\frac{5}{4}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 9}{4}=\frac{9}{2}$
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