Given:
The equation is \( \dfrac{y^2}{9} – \dfrac{x^2}{27} = 1\)
or \( \dfrac{y^2}{3^2} – \dfrac{x^2}{27^2} = 1\)
On comparing this equation with the standard equation of hyperbola \( \dfrac{ y^2}{a^2} – \dfrac{x^2}{b^2} = 1,\)
We get a = 3 and b = \( \sqrt{27}\),
It is known that, a2 + b2 = c2
So,
c2 = \( 3^2+( \sqrt{27})^2\)
= 9 + 27
c2 = 36
c = \( \sqrt{36}\)
= 6
Then,
The coordinates of the foci are (0, 6) and (0, -6).
The coordinates of the vertices are (0, 3) and (0, – 3).
Eccentricity, e = \( \dfrac{c}{a}= \dfrac{6}{3}=2\)
Length of latus rectum = \( \dfrac{2b^2}{a}\)
= \(\dfrac{( 2 × 27)}{3} = \dfrac{(54)}{3} = 18\)
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