# Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas : $5y^2-9x^2=36$

$\begin {array} {1 1} (A)\;Foci : ( 0, \pm 2\sqrt{14} / \sqrt{5} ), Vertices : ( 0, \pm 6 / \sqrt 5), e = \sqrt{14} / 3, Length \; of \: the \: latus\: rectum 4\sqrt 5 / 3 \\ (B)\;Foci : ( \pm 2\sqrt{14} / \sqrt{5} , 0), Vertices : ( \pm 6 / \sqrt 5, 0), e = \sqrt{14} / 3, Length \; of \: the \: latus\: rectum 4\sqrt 5 / 3 \\ (C)\;Foci : ( 2\sqrt{14} / 5, 0 ), Vertices : ( 6 / \sqrt 5, 0), e = \sqrt{14} / 3, Length \; of \: the \: latus\: rectum 4\sqrt 5 / 3 \\ (D)\;Foci : ( -\pm 2\sqrt{14} / \sqrt{5}, 0 ), Vertices : ( -6 / \sqrt 5, 0), e = \sqrt{14} / 3, Length \; of \: the \: latus\: rectum 4\sqrt 5 / 3 \end {array}$

Toolbox:
• Standard equation of a hyperbola where transverse axis is along x - axis is $\large\frac{x^2}{a^2}$$-\large\frac{y^2}{b^2}$$=1$
• Standard equation of a hyperbola whose transverse axis is along y - axis is $\large\frac{y^2}{a^2} $$- \large\frac{x^2}{b^2}$$=1$
• $c^2=a^2+b^2$, where $c$ is the foci
• Eccentricity $e = \large\frac{c}{a}$
• Length of the latus rectum $= \large\frac{2b^2}{a}$
Step 1 :
Given equation is $5y^2-9x^2=36$
Dividing throughout by 36 we get,
$\large\frac{y^2}{\Large\frac{36}{5}} $$- \large\frac{x^2}{4}$$=1$
Comparing this equation with the standard equation of the hyperbola
$\large\frac{y^2}{a^2}$$- \large\frac{x^2}{b^2}$$=1$ we get,
$a^2= \large\frac{36}{5}$ and $b^2=4$
$\therefore c^2 = a^2+b^2$
(i.e) $c^2= \large\frac{36}{5}$$+4 = \large\frac{56}{5} \therefore c = \sqrt{ \large\frac{56}{5}}$$ = \large\frac{2\sqrt{14}}{\sqrt 5}$
Step 2 :
Eccentricity $e = \large\frac{c}{a}$$=\large\frac{\sqrt{ \large\frac{56}{5}}}{\large\frac{6}{\sqrt 5}} = \large\frac{\sqrt{14}}{3} Coordinates of foci are \bigg(0, \pm \large\frac{2\sqrt{14}}{\sqrt 5} \bigg) Coordinates of vertices are \bigg(0, \pm \large\frac{6}{\sqrt 5} \bigg) Length of the latus rectum = \large\frac{2b^2}{a}$$ = \large\frac{2 \times 4 }{\Large\frac{6}{\sqrt 5}}$
$= \large\frac{4\sqrt 5 }{3}$