Question

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

Answer 1

The given equation is $\dfrac{y^2}{9} – \dfrac{x^2}{27} = 1$
On comparing this equation with the standard equation of hyperbola i.e., $\dfrac{y^2}{a^2} – \dfrac{x^2}{b^2} = 1$

We obtain $a = 3$ and $b = √27$.
We know that
$a^2 + b^2 = c^2 .$
$∴ c^2 = 3^2 \+ (√27)^2$
$= 9 + 27$
$c^2 = 36$
$c = √36$
$= 6$

Therefore, The coordinates of the foci are $(0, ±6).$
The coordinates of the vertices are $( 0,±3).$
Eccentricity, $e = \dfrac{c}{a} = \dfrac{6}{3} = 2$
Length of the latus rectum $= \dfrac{2b^2}{a} = \dfrac{(2 × 27)}{3}$

$= \dfrac{(54)}{3} =18$