Given equation of hyperbola is 5y2–9x2 = 36
i.e.,

5y2

36

−

9x2

36

= 1 ⇒

y2

36 5

−

x2

4

= 1,
which is of the form

y2

a2

−

x2

b2

= 1.
The foci and vertices of the hyperbola lie on y-axis.
Now, a2 =

36

5

⇒ a =

6

√5

and b2 = 4 ⇒ b = 2
Also, x2 = a2+b2 =

36

5

+4 =

56

5

⇒ c = √

56

5

= 2√

14

5

∴ Coordinates of foci are (0, ± c) i.e. (0,±

2√14

√5

)
Coordinates of vertices are (0, ± a) i.e. (0,±

6

√5

)
Eccentricity (e) =

c

a

=

2√ 14 5

6 √5

=

√14

3

Length of latus rectum =

2b2

a

=

2×4

6 √5

=

4√5

3

.