Given equation of hyperbola is 49y2–16x2 = 784
i.e.,

49y2

784

−

16x2

784

= 1 ⇒

y2

16

−

x2

49

= 1
which is of the form

y2

a2

−

x2

b2

= 1.
The foci and vertices of the hyperbola lie on y-axis.
Now, a2 = 16 ⇒ a = 4 and b2 = 49 ⇒ b = 7
Also, c2 = a2+b2 = 16 + 49 = 65 ⇒ c = 65
∴ Coordinates of foci are (0, ±c) i.e. (0,±√64)
Coordinates of vertices are (0, ±a) i.e. (0, ±4)
Eccentricity (e) =

c

a

=

√65

4

Length of latus rectum =

2b2

a

=

2×49

4

=

49

2

.