Question

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola  \(9y^2-4x^2=37\)

Answer 1

The given equation is \(9y^2 - 4x^2 = 36.\)
or \(\dfrac{y^2}{4} – \dfrac{x^2}{9} = 1 \)

or \(\dfrac{y^2}{2^2} – \dfrac{x^2}{3^2} = 1.......(1)\)
On comparing equation (1) with the standard equation of hyperbola i.e., \(\dfrac{y^2}{a^2} – \dfrac{x^2}{b^2} = 1 \) we obtain \(a = 2\) and \(b = 3\). \(\)
We know that \(a^2 + b^2 = c^2 .\)
\(∴ c^2 = 4 + 9\)
\(c^2 = 13\)
\(c = √13.\)

Therefore, 
The coordinates of the foci are \((0, √13)\) and \((0, –√13).\)
The coordinates of the vertices are \((0, 2)\) and \((0, – 2).\)
Eccentricity, \(e = \dfrac{c}{a} = \dfrac{√13}{2}\)
Length of the latus rectum \(= \dfrac{2b^2}{a} = \dfrac{(2 × 3^2)}{2} \)

\(= \dfrac{(2×9)}{2} = \dfrac{18}{2} = 9\)