Question

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

Answer 1

The given equation is $\dfrac{x^2}{16} – \dfrac{y^2}{9} = 1$ or $\dfrac{x^2}{4^2} –\dfrac{y^2}{3^2} = 1$
On comparing this equation with the standard equation of hyperbola i.e., $\dfrac{x^2}{a^2} – \dfrac{y^2}{b^2} = 1$, we obtain $a = 4$ and $b = 3.$
We know that
$a ^2 + b ^2 = c^2$
$∴ c^2 = 4^2 \+ 3^2$
$= √25$
$c = 5$

Therefore, The coordinates of the foci are $( ±5, 0).$
The coordinates of the vertices are $( ±4, 0).$
Eccentricity, $e = \dfrac{c}{a} = \dfrac{5}{4}$
Length of the latus rectum $= \dfrac{2b^2}{a} = \dfrac{(2 × 32)}{4}$

$= \dfrac{(2×9)}{4}$

$= \dfrac{18}{4}$

$= \dfrac{9}{2}$