UPSC NDA Math Model Paper 4

To find the foci of hyperbola first find eccentricity.
Given , equation of hyperbola
4x2−9y2−1=0
⇒4x2−9y2=1
⇒

x2

1 4

−

y2

1 9

Here, a2=

1

4

and b2=

1

9

⇒a2>b2
The Eccentricity of the hyperbola is e=√1+

b2

a2

⇒e=√1+

(1∕9)

(1∕4)

⇒e=√1+

4

9

⇒e=

√13

3

The foci of hyperbola =(±ae,0)
Here a=

1

2

Foci =(±

1

2

√13

3

,0)
Hence, the Foci of hyperbola =(±

√13

6

,0)