Question

3. 9y2_ 4x2-36

Answer 1

The given equation of the hyperbola is 9 y 2 −4 x 2 =36 .

9 y 2 −4 x 2 =36 (1)

The above equation can be written as,

y 2 36 9 − x 2 36 4 =1 y 2 4 − x 2 9 =1

Since the transverse axis is along the y axis, equation of the hyperbola can be represented as y 2 a 2 − x 2 b 2 =1 .(2)

Comparing equations (1) and (2), we get

a=2 and b=3

Now, we know c 2 = a 2 + b 2 c 2 = 2 2 + 3 2 c 2 =4+9 c 2 =13 c= 13

Since y axis is the transverse axis, coordinates of Foci = (0,±c)=(0,± 13 )

Since y axis is the transverse axis, coordinates of Vertices = (0,±a)=(0,±2)

Eccentricity = e = c a = 13 2

Length of Latus rectum = 2 b 2 a = 2 (3) 2 2 =9

Thus, the coordinates of foci of hyperbola 9 y 2 −4 x 2 =36 are ( 0,± 13 ) , coordinates of vertices are ( 0,±2 ) , eccentricity is 13 2 and length of latus rectum is 9 .