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Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $\;\large\frac{x^2}{16}$$-\large\frac{y^2}{9}$$=1$

1 Answer

Toolbox:
On comparing the given equation with the standard equation of a hyperbola, we get:
$\quad q = 4, b = 3$
Since $a^2+b^2 = c^2 \rightarrow c = \sqrt{16+9} = \sqrt {25} = 5$
1. Coordinates of Foci $ = (\pm 5, 0)$
2. Coordinates of Vertices $= (\pm 4, 0)$
3. Eccentricity $e = \large\frac{c}{a} $$=\large\frac{5}{4}$
4. Latus Rectum $=\large\frac{2b^2}{a}$$ = \large\frac{2\times3^2}{4} $$= \large \frac {9}{2}$
answered Apr 8, 2014 by balaji.thirumalai
 

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